Chapter 01 Electrostatics Part 1 With Watermark

.td .L vt aP di Electrostatics: In Part 1 g in rna Le a ge ng Ce Module-Vol-II_01.indd 1 3/31/2017 4:31:24 PM . td .L vt aP di In g in a rn Le a ge ng © 2017 Cengage Learning India Pvt. Ltd. Ce ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmitted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, without the prior written permission of the publisher. Module-Vol-II_01.indd 2 3/31/2017 4:31:24 PM . td Electrostatics: Part 1 .L vt aP di In 1.  Electric Charge The electrical nature of matter is inherent in atomic structure. An atom consists of a small, relatively massive nucleus that contains particles called protons and neutrons. A proton has a mass of 1.673 × 10 −27 kg, and a neutron has a slightly greater mass of g 1.675 × 10−27 kg. Surrounding the nucleus is a diffuse cloud of orbiting particles called electrons, as the figure suggests. An electron has a mass of 9.11 × 10−31 kg. Like mass, ‘electric charge’ is in an intrinsic property of protons and electrons, and only two types of charge have been discovered, positive and negative. A proton has a positive charge, and an electron has a negative charge. A neutron has no net electric charge. rn Experiment reveals that the magnitude of the charge on the proton exactly equals the magnitude of the charge on the electron; the proton carries a charge +e, and the electron carries a charge −e. The SI unit for measuring the magnitude of an electric charge is the coulomb1 (C), a and e has been determined experimentally to have the value, e = 1.60 × 10–19 C. (i) Number of protons = number of electrons. Le (ii) Protons have the basic +e charge and electrons have the basic –e charge. (iii) Hence a normal atom is electrically neutral. Electrons can travel from one atom to another and from one body to another. ge If a body loses one electron, it becomes positively charged with +e charge and vice-versa. 2. Charging of a Body a Ordinarily, matter contains equal number of protons and electrons. A body can be charged by the transfer of electrons or redistribution of electrons. Basically charging can be done by three methods: ng (i) Friction (ii) Conduction (iii) Induction By friction: In friction when two bodies are rubbed together, electrons are transferred from one body to the other. As a result of this one body becomes positively charged while the other negatively charged, e.g., when a glass rod is rubbed with silk, the rod Ce becomes positively charged while the silk negatively. However, ebonite on rubbing with wool becomes negatively charged making the wool positively charged. Clouds also become charged by friction. In charging by friction in accordance with Module-Vol-II_01.indd 1 3/31/2017 4:31:25 PM 1  Properties of Electric Charge (i) Charge is transferable: If a charged body is put in contact with an uncharged body. the charged body will attract opposite charge and repel similar charge present in the uncharged body.indd 2 3/31/2017 4:31:26 PM . both positive and negative charges in equal amounts appear td simultaneously due to transfer of electrons from one body to the other. If the grounding wire is then removed. In the In figure negatively charged rod is brought close to. As a result of this one side of neutral body (closer to charged body) di becomes oppositely charged while the other is similarly charged. The process of giving one object a net electric charge without touching the object to a second charged object is called ‘charging a by induction’. the part of the sphere nearest the rod becomes positively charged and the part farthest away becomes negatively charged.. as in the figure. the sphere is left with a positive net charge. if a positively charged rod were used. as part (a) of the drawing indicates. Then. the sphere would be left with a positive charge after being rubbed with a positively charged rod. This process is called electrostatic induction. Le a ge ng 2. such as the sphere in Figure (a). 2  Electrostatics: Part 1 . aP The process of giving one object a net electric charge by placing it in contact with another object that is already charged is known as ‘charging by contact’. some of the excess electrons from the . (ii) Charge is always associated with mass: i. they repel one another and spread out over the sphere’s surface. where they could spread out even more. It is also possible to charge a conductor in a way that does not involve contact. Module-Vol-II_01. In the sphere. but does not touch. uncharged body becomes charged due Ce to transfer of electrons from one body to the other. charge cannot exist without mass though mass can exist without charge. electrons from the sphere would be transferred to the rod. the free electrons closest to the rod move to the other side. the vt sphere is left with a negative charge distributed over its surface. In a similar manner. As a result. As shown in Figure (b) of the picture.L rod are transferred to the object. as part (c) of the picture shows.e. electrons would be drawn up from the ground through the grounding wire and onto the sphere. and the charged regions would disappear. By electrostatic induction: If a charged body is brought near an uncharged body. some of the free electrons leave the sphere and distribute themselves over the much larger earth. Charging by conduction: When a negatively charged ebonite rod is rubbed on a metal object. followed by the ebonite rod. If the rod in were removed. rn Under most conditions the earth is a good electrical conductor. These positively and g negatively charged regions have been “induced” or “persuaded” to form because of the repulsive force between the negative rod and the free electrons in the sphere. In this case. The process could also be used to give the sphere a negative net charge. So when a metal wire is attached between the sphere and the ground. conservation of charge. the free electrons would return to their original places. The insulated stand prevents them from flowing to the earth. Once the electrons are on the metal sphere (where they can move readily) and the rod is removed. a metal sphere. e. as quarks do not exist in free state.g.25 ¥ 1018 electronic charges. 3 rn Note: • Recently it has been discovered that elementary particles such as proton or neutron are composed of quarks having charge (± 1/3)e and (± 2/3) e. If the charge of an electron ( 1. If the charges were dependent on velocity. i. And if the motion of charged particle is accelerated it not only produces electric and magnetic fields but also radiates energy in the aP space surrounding the charge in the form of electromagnetic waves. di (vi) Charge resides on the surface of conductor: Charge resides on the outer surface of a conductor because like charges repel and try to get as far away as possible from one another and stay at the farthest distance from each other which is outer surface In of the conductor. The smallest charge that can exist in nature is the charge of an electron. Why? What happens if the hair is wet or it is rainy day? Solution ge When the comb runs through dry hair. This is why a solid and hollow conducting sphere of same outer radius will hold maximum equal charge and a soap bubble expands on charging.6 ¥ 10 -19 this of large number. (viii) Quantization of charge: When a physical quantity can have only discrete values rather than any value. e. large-scale charges? ng Solution 1 Large-scale charges contain a very large number of electrons. (iii) Charge is conserved: Charge can neither be created nor be destroyed.e.indd 3 3/31/2017 4:31:26 PM . However. it gets charged due to friction. 2. a  Illustration 2    Why can one ignore quantization of electric charge when dealing with macroscopic. For example. Explain how this observation is consistent with the law of conservation of charge. On account 1. Q = ± ne with n = 1. Electrostatics: Part 1  3 . However. vt (v) Charge produces electric field and magnetic field: A charged particle at rest produces only electric field in the space surrounding it. there is no friction between the hair and the comb and hence the comb remains uncharged. 1C = = 6. if the charged particle is in unaccelerated motion it produces both electric and magnetic fields.. the neutrality of atoms would be violated. a • Quantization of charge implies that there is a maximum permissible magnitude of charge. these acquires induced charges whose nature is opposite to that on the comb. When this comb is brought near small bits of paper.6 × 10–19C) g is taken as elementary unit i. The difference in masses on an electron and a proton suggests that electrons move much faster in an atom than protons. But on a rainy day or when the hair is wet. charge flows continuously and the quantization of charge can be ignored. the bits of paper get attached to the comb.e. the quantity is said to be quantised. 3  in 2 Charge on a body can never be ± e. [NCERT] Module-Vol-II_01. the quanta of charge is still e. charges appear on both. Ce  Illustration 3    When a glass rod is rubbed with a silk cloth..L Thus the total charge is +92e both before and after the decay.2 e or ± 10 -5 e etc. (iv) Invariance of charge: The numerical value of an elementary charge is independent of velocity. ± 17. Le  Illustration 1    A comb runs through one’s dry hair and attracts small bits of paper. quanta of charge the charge on anybody will be some integral multiple of e i.. Consequently. It is proved by the fact that an atom is neutral. A similar phenomenon is observed with many other pairs of bodies. in radioactive decay the uranium nucleus td (charge = +92e) is converted into a thorium nucleus (charge = +90e) and emits an a-particle (charge = +2e) 238 92 U Æ90 Th 234 + 2 He 4 . L  Illustration 4    A glass rod is rubbed with a silk cloth. Originally A and B both have charges a of +3 mC.  Illustration 6    Objects A. Find the number of electrons lost by glass rod. [Neglect effect due to induction]. Solution a When ‘A’ and ‘B’ are rubbed. Solution td The positive charge developed on the glass rod is found to have the same magnitude as the negative charge developed on silk cloth. If. This is expected because the mass of an electron is extremely small. Le • Electrons are loosely bound with ‘A’ in comparison to ‘B’ . me = 9 × 1031 kg aP Solution q 19. they will (a) attract (b) repel (c) have no effect on each other. (d) B becomes negatively charged and C becomes negatively charged. ‘A’ becomes positively charged and ‘B’ becomes negatively charged. there will be transfer of mass from glass rod to silk cloth.indd 4 3/31/2017 4:31:26 PM . they will (a) attract (b) repel (c) have no effect on each other. di Mass transferred = 12 ¥ (9 ¥ 10 -31 ) = 1. When ‘A’ and ‘C’ are rubbed together. (A) If objects A and B are now held near each other.6 ¥ 10 -19 (i) Charge on silk = −19.08 ¥ 10 -29 kg Note that mass transferred is negligibly small. then g A becomes negatively charged. Module-Vol-II_01.2 ¥ 19 -19 Number of electrons lost by glass rod is n = = = 12 e 1. (b) B becomes positively charged and C becomes negatively charged. then A becomes positively charged and B becomes negatively charged. ge • Hence if the object ‘B’ and ‘C’ are rubbed together ‘C’ will lose electrons and ‘B’ will receive electrons. In  Illustration 5    If an object made of substance A rubs an object made of substance B. and they are moved apart. insulated. an object made of substance A is rubbed against an object made of substance C. which is consistent with the law of conservation of charge. • Hence in ‘C’ electrons are most loosely bound. ‘A’ becomes negatively charged. B and C are three identical. Objects A and C are allowed to touch. • Hence ‘C’ will become positively charge and ‘B’ will become negatively charged. while C has a charge of –6 mC. • Electrons are loosely bound with ‘C’ in comparison to ‘A’. Then objects B and ng C are allowed to touch.2 × 10–19 C. What will happen if an object made of substance B is rubbed against an object made of substance C? in (a) B becomes positively charged and C becomes positively charged. Ce (B) If instead objects A and C are held near each other. As such. . Find the negative charge acquired by silk. then they are moved apart. 4  Electrostatics: Part 1 . vt Is there transfer of mass from glass to silk? Given. rn (c) B becomes negatively charged and C becomes positively charged. however. The glass rod acquires a charge of + 19. the total charge on the glass rod and the silk cloth is the same before and after rubbing. spherical conductors.2 × 10–19 C (ii) Since an electron has a finite mass (me = 9 × 10–31 kg). indd 5 3/31/2017 4:31:27 PM . Both the spheres will become uncharged. ∑ If A and C are now held near each other they will also attract each other. This movement leaves unbalanced positive charge on B. In  Illustration 7    Consider figure. and same move in the left side of A. insulated stands and placed in contact with each other. then move apart vt ∑ When the objects B and C are allowed to touch.L ∑ When the objects A and C are allowed to touch. They will remain so even if they have different size or material. the free electrons in the sphere A are attracted to the rod. If the rod is removed before the spheres are separated. a ng Ce Module-Vol-II_01. the excess electrons on sphere A would flow back to B. a positively charged rod is brought near two uncharged metal spheres A and B attached with What would happen if the rod were removed before the spheres are separated? g Would the induced charges be equal in magnitude even if the spheres had different sizes and different conductors? in What will happen when the rod is removed after the spheres are separated? Solution rn (a) When a positively charged rod is brought near A. Before the rod is brought near A. both A and B were neutral. a Le ge (b) Yes. Electrostatics: Part 1  5 . net charge is conserved. aP di Hence if A and B are now held near each other they will attract each other. then move apart. Solution td . 6 × 10–19 c (c) 1.6 × 10–10 c (d) 4.6 × 10–17 c (b) 1.8 × 10–10 c 2. When metal knob is touched with a charged body. the sphere A will have net negative charge and sphere B td will have net positive charge of same magnitude. which then diverges due to repulsion.L It is a simple apparatus with which the presence of electric charge on a body is detected (see figure). Which one of the following statements regarding electrostatics is wrong? a (a) Charge is quantized ng (b) Charge is conserved (c) There is an electric field near an isolated charge at rest (d) A stationary charge produces both electric and magnetic fields Ce 3. If a charged body brought near a charged electroscope. 6  Electrostatics: Part 1 . the leaves will further diverge. One quantum of charge should be at least be equal to the charge in coulomb: ge (a) 1. Select the correct alternative/ alternatives: (a) The charge gained by the uncharged body from a charged body due to conduction is equal to half of the total charge initially present Module-Vol-II_01. (c) As charge is conserved if rod is removed and spheres are separated. If the induction effect is strong enough leaves after converging may again diverge. The separation gives a rough idea of the amount of charge on the body.  Electroscope . If vt the charge on body is similar to that on electroscope and will usually converge if opposite. some charge is transferred to the gold leaves.indd 6 3/31/2017 4:31:27 PM . 3. (i) Uncharged electroscope aP di In g (ii) Charged electroscope in a rn Le Concept Application Exercise 1 1. In the return stroke of a typical lightning bolt. each object is ge repelled from the other. If. How many megacoulombs of positive charge are in 1. figure shows two charged bodies. each one acting on a different object. Coulomb’s Law a The electrostatic force that stationary charged objects exert on each other depends on the amount of charge on the objects and the Le distance between them. an oxygen atom contains eight protons. Next. with a g value given by r = bx2. as in part (b). These objects are so small. Calculate the number of coulombs of positive charge in 250 cm3 of (neutral) water.80 A in its filament.0 × 10−31 kg) 11. with a length of 2.0 × 1023) vt 6. Electrons and positrons are produced by the nuclear transformations of protons and neutrons known as beta decay. and (b) non-uniform. These forces always exist as a pair. as in part (a) of the picture. The repulsive forces. in accord with Newton’s action–reaction law. The “point charges” have magnitudes |q1| and |q2|. The volume charge density r is charge per unit volume in coulombs per cubic meter. sphere W is touched to sphere B (with an initial charge of −32e) and then they are separated.0 kg of electrons? (The mass of an electron = 9. sphere W is touched to sphere C (with an initial charge of +48e). What was the initial charge on sphere A? 9. A 100 W lamp has a steady current of 0. + F is the electric force exerted on object 1 by object 2 and . If the charges have unlike signs. whose experiments in 1785 led him to it. Finally. td (b) The magnitude of charge increases with the increase in velocity of charge (c) Charge cannot exist without matter although matter can exist without charge (d) Between two non-magnetic substances repulsion is the true test of electrification (electrification means body has net . How many excess electrons are on the rod if r is (a) uniform. lies along the positive side of an x axis with one end at the origin. act along the line between the charges. (Hint: A hydrogen atom contains one proton. How long is required for 1 mol of electrons to pass through rn the lamp? (NA= 6. where particle 1 has charge q1 and particle 2 has charge q2. and then they are separated. is an electron or a positron produced? (b) If a neutron transforms into a proton.) aP 7.F is the electric force exerted on object 2 by object 1.00 m and a cross-sectional area of 4. A charged non-conducting rod. a +F –F –F q1 q2 +F q1 + – q2 + + ng r r (a)       (b) Ce The equation for the electrostatic forces acting on the particles is called Coulomb’s law after Charles-Augustin de Coulomb.5 × 104 A exists for 20 ms.indd 7 3/31/2017 4:31:27 PM . (These symbols can represent either positive or negative charge. like the attractive forces.00 µC/m3.00 µC/m5? in 10. The In final charge on sphere W is +18e. How much charge is transferred in this event? 5. where b = −2. the charges have the same sign (both positive or both negative). What is the total charge in coulombs of 72. that they can be regarded as mathematical points.L charge) 4. Electrostatics: Part 1  7 . (a) If a proton transforms into a neutron.) Let’s also focus on particle 1 and write the force acting on it in terms of a unit vector r that points along a radial axis extending through the Module-Vol-II_01. compared to the distance r between them.00 mol of neutral molecular-hydrogen gas (H2)? (NA = 6.00 cm2. each object is attracted to the other by a force   that is directed along the line between them. Whether attractive or repulsive. with a value of −4. Let’s write the equation in vector form and in terms of the particles shown in figure. Figure shows four identical conducting spheres that are actually well separated from one another. a current of 2. To set the stage for explaining these features in more detail. Sphere W (with an initial charge of zero) is touched to sphere di A and then they are separated. the two forces are equal in magnitude but opposite in direction. is an electron or a positron produced? 8.0 × 1023) 4. rn 4.. two particles.  aP . F12 = F21 = 4pe 0 r 2 Module-Vol-II_01. For all other media it is greater than one and for conductors it is infinite. r̂21 be the unit vector from q2 to q1.) If between the two charges there is free space then. That checks because particle 1 is being attracted toward particle 2.  Le Let F12 be the force on charge q1 due to q2. the magnitude of forces are equal i. Eq. Due to like nature of charges. they a will repel each other. (i). Next. if q1 and q2 have opposite signs. r̂12 be the unit vector from q1 to q2.   1 q1q2 It is also clear from above equations. now Eq. di 1 k= = 9 ¥ 109 Nm 2 C -2 (in SI Units) 4pe 0 In where e0 = (8. F12 and rˆ21 are in same direction  1 q1q2 \ F12 = rˆ 4pe 0 r 2 21 Ce The above equations give the Coulomb’s law in vector form. That checks. then the product q1q2 in gives us a positive result.) With these decisions.85 × 1022 C2N–1m–2) is the absolute electric permittivity of the free space. it is clear that F21 and rˆ12 and are in the same direction  1 q1q2 \ F21 = rˆ a 4pe 0 r 2 12  ng Also. So.L vt  qq F = k 1 2 2 r (Coulomb's law).e. we write the electrostatic force as . r has a magnitude of exactly 1 and no unit.(i) r where r is the separation between the particles and k is a positive constant called the electrostatic constant or the Coulomb constant.. If q1 and q2 have the same sign. (i) tells us that the force on particle 1 is in the direction of r.. separated by a distance r. ge  From figure.1  Coulomb’s Law in Vector Form Let q1 and q2 be two like charges placed at points A and B in vacuum. (As with other unit vectors. radially away from particle 2.indd 8 3/31/2017 4:31:28 PM . like a direction arrow on a street sign. 8  Electrostatics: Part 1 . the product q1q2 gives us a negative result.  F21 be the force on charge q2 due to q1. its purpose is td to point. So. because particle 1 is being repelled from particle 2. (We’ll discuss k below. Let’s first check the direction of the force on particle 1 as given by Eq. (i) tells us that the force on particle 1 is in the direction opposite r. If there is medium between the two charges 1 then k = where er is the relative permittivity of medium which is also known as the dielectric constant. Relative permittivity 4pe 0e r g (er ) of vacuum is one.   1 q1q2          ∵ r12 aP or F21 = r r12 = ˆ 4pe 0 r 3 12 r12 12 Similarly. Q3 … are applying force on a charge Q Net force on Q will be Le      Fnet = F 1+ F2 +  + Fn -1 + Fn  Illustration 8    What is the force between two small charged spheres having charges of 2 × 10–7 C and 3 × 10–7 C placed 30 cm apart in air?  [NCERT] ge Solution Ê Nm 2 ˆ (2 ¥ 10 -7 C) (3 ¥ 10 -7 C) F = 9 ¥ 109 Á 2 ˜ -2 2 = 6 ¥ 10 -3 N (repulsive) (30 ¥ 10 m) a Ë C ¯ ng  Illustration 9    The electrostatic force on a small sphere of charge 0.e. opposite to each other i. Therefore. vt Also the forces due to two point charges are parallel to the line joining the point charges. F12 = (– ˆ r12 ). Q2.8 µC in air is 0. total force acting rn on a given charge due to number of charges is the vector sum of the individual forces acting on that charge due to all the charges. Principle of superposition: According to the principle of super position. F12 = – F21 Thus. rˆ21 = – rˆ12 td  1 q1q2  1 q1q2 Put this value in equation. a Consider number of charge Q1.. (a) What is the distance between the two spheres? Ce (b) What is the force on the second sphere due to the first? [NCERT] Module-Vol-II_01. such forces are called central forces and so electrostatic forces are conservative forces. Since rˆ12 and rˆ21 are unit vectors.2 N.indd 9 3/31/2017 4:31:29 PM . g Note: in Coulomb’s law is true for point charges only and the force between any two charges is not affected by the presence of other charges. Newton’s third law is obeyed.L From equations. the forces exerted by two charges on each other are equal in magnitude and opposite in direction. Electrostatics: Part 1  9 .4 µC due to another small sphere of charge –0. force on q1 due to q2 is  1 q1q2  1 q1q2  F12 =  ˆ r or F12 = r 4pe 0 r 3 21 di 21 4pe 0 r 2 21 21 The above equations can also be written in the following manner: In  1 q1q2    1 q1q2   F21 =   (r – r1 ) and F12 = (r – r ) 4pe 0 r – r 3 2 4pe 0 r – r 3 1 2  2 1 1 2 These equations represent Coulomb’s law in terms of position vectors. F12 = – rˆ   4pe 0 r 2 4pe 0 r 2 12 . 10  Electrostatics: Part 1 . Solution td kq1q2 kq1q2 (a) As F = , r= 2 r F .L 1/2 Ê 9 ¥ 109 (Nm 2 /C2 ) (0.4 ¥ 10 -6 C) (0.8 ¥ 10 -6 C) ˆ r=Á ˜ = 144 ¥ 10 -4 m 2 = 12 ¥ 10 -2 m = 12 cm Ë 0.2 N ¯ vt (b) Since the two charges are of opposite signs, the force between them is attractive in nature. Further, according to Newton’s third law, the force on the second sphere due to first is also attractive and has a magnitude 2 N. aP  Illustration 10    (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation. di (b) What is the force of repulsion if: (i) Each sphere is charged double the above amount, and the distance between them is halved. (ii) The two spheres are placed in water? (Dielectric constant of water = 80) [NCERT] Solution (a) We are given that, q1 = q2 = 6.5 × 10–7 C, r = 50 cm = 0.5 m If F is the force of repulsion, F = ke q1 q2 In g r2 in (6.5 ¥ 10 -7 ) (6.5 ¥ 10 -7 ) or F = (9 ¥ 109 ) ¥ 2 = 1.5 ¥ 10 -2 N. (0.5) rn (b) (i) If the charge on each sphere is doubled and the distance between them is halved, the force of repulsion, F´ would become 16 times (as F µ q1q2 and F µ 1/r2).   Clearly, F´ = 16F = 16 (1.5 × 10–2 N) = 0.24 N a F 1.5 ¥ 10 -2 (ii)  If the sphere are placed in water (er = 80), Fw = = = 1.9 ¥ 10 -4 N Œr 80 Le  Illustration 11    Suppose the spheres A and B in previous illustration have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then in contact with the second, and finally removed from both. What is the new ge force of repulsion between A and B? Solution We are given that, Charge on sphere A = charge on sphere B = q = 6.5 × 10–7 C Force of repulsion between A and B, i.e., a q¥q q2 (6.5 ¥ 10 -7 )2 F = ke = ke = (9 ¥ 109 ) = 1.5 ¥ 10 -2 N ng 2 2 2 r r (0.5) Let the third sphere (say C) of the same size but uncharged be brought in contact with A. Due to the flow of electrons, the two spheres share the charge equally. Therefore, Ce q+0 q Charge on A = Charge on C = = 2 2 When the sphere C (having charge q/2) touches the sphere B (having charge q), Module-Vol-II_01.indd 10 3/31/2017 4:31:29 PM Electrostatics: Part 1  11 . Total charge on B and C (q + q/2 = 3q/2) is equally distributed between them. td Thus, charge on B = charge on C = 3q/4 If F´ is the new force of repulsion between A and B, .L (q / 2) (3q / 4) 3 q2 3 3 F ¢ = ke = ke = F = (1.5 ¥ 10 -2 N) = 5.7 ¥ 10 -3 N r2 8 r 2 8 8 vt  Illustration 12    Two particles A and B having charges 8 × 10–6 C and –2 × 10–6C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force? aP Solution As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC = x and the charge on C is Q di  FCA = 1 (8.0 ¥ 10 -6 ) Q ˆ i In g 4p Œ0 (0.2 + x )2  -1 (2.0 ¥ 10 -6 )Q ˆ in and FCB = i 4p Œ0 x2    È (8.0 ¥ 10 -6 ) Q (2.0 ¥ 10 -6 ) Q ˘ rn 1 FC = FCA + FCB = Í - ˙i 4p Œ0 ÍÎ (0.2 + x ) 2 x2 ˙˚ But  | FC | = 0 a È (8.0 ¥ 10 -6 ) Q (2.0 ¥ 10 -6 ) Q ˘ Le 1 Hence Í - ˙ = 0 which gives x = 0.2 m 4p Œ0 ÎÍ (0.2 + x ) 2 x2 ˚˙  Illustration 13    Point charges Q and q are placed at the vertices of a square of side a as shown. What should be sign of charge ge q q and magnitude of ratio of so that Q (a) net force on each Q is zero a (b) net force on each q is zero Is it possible that the entire system could be in electrostatic equilibrium? ng Solution Ce Case I: Let the charge q and Q are of same sign. (a) Consider the forces acting on charge Q placed at A (shown in figures (A) and (B)) Module-Vol-II_01.indd 11 3/31/2017 4:31:30 PM 12  Electrostatics: Part 1 . td .L vt aP 1 qQ Here F1 = {force of q at D on Q at A} 4pe 0 a 2 di 1 qQ F2 = {force of q at B on Q at A} 4pe 0 a 2 In 1 QQ F3 = {force of Q at C on Q at A} 4pe 0 2a 2    In Figure (A), resultant of forces F1 and F2 will lie along F3 so that net force on Q cannot be zero. Hence, q and Q have to g be of opposite signs. Case II: Let the charge q and Q are of opposite sign. in    In this case, as shown in Figure (B), resultant of F1 and F2 will be opposite to F3 so that it becomes possible to obtain a condition of zero net force. rn    Let us write FR = F1 + F2 1 qQ \ Fr = F12 + F22 = 2 a 4pe 0 a 2   Le Direction of FR will be along AC ( FR , being resultant of forces of equal magnitude, bisects the angle between the two)   FR and F3 are in opposite directions. Net force on Q can be zero if their magnitudes are also equal, i.e., 1 qQ 1 QQ Q Ê Qˆ 2= or Á 2 q - ˜¯ = 0 4pe 0 a 2 Ë ge 4pe 0 a 2 4pe 0 2a 2 2 Q q 1 ⇒ q= fi = {Q π 0} 2 2 Q 2 2 a The sign of q should be negative. ng (b) Consider now the forces acting on charge q placed at B. In a similar manner, as discussed in (a), for net force on q to be zero, q and Q have to be of opposite signs. This is also shown in the given figures. Ce Module-Vol-II_01.indd 12 3/31/2017 4:31:30 PM 1 Qq 1 q2 2= a 4pe 0 a 2 4pe 0 2a 2 Le q Ê qˆ q q \ Á 2Q . for equilibrium of a particle along a vertical line. Module-Vol-II_01. F3 is opposite to F3 . 2 2 2 2 ng  Illustration 14    Three particles.L vt aP 1 Qq Now. td . are suspended from a common point by insulating massless strings.. If the particles are in equilibrium and are located at the corners of an equilateral triangle of side Ce ‘a’.. q = . Net force can become zero if their magnitudes are also equal i. calculate the charge q on each particle. each of mass ‘m’ and carrying a charge q..e. i. Electrostatics: Part 1  13 . F1 = {force of Q at A on q at B} 4pe 0 a 2 di 1 Qq F2 = {force of Q at C on q at B} 4pe 0 a 2 In 1 q2 F3 = {force of q at D on q at B} 4pe 0 2a 2 Referring to Figure (D) g    Let us write FR = F1 + F2 in \ 1 Qq FR = F12 + F22 = 2 4pe 0 a 2    rn  Resultant of F1 and F2 . each ‘L’ long. Solution In accordance with (shown in Figure (B)).e.e. cannot be satisfied together. ˜¯ = 0 fi Q = fi =2 2 {q π 0} 4pe 0 a Ë 2 2 2 2 Q The sign of ‘q’ should be negative.indd 13 3/31/2017 4:31:31 PM . and Q = . i. In this case we need not to repeat the calculation as the present situation is same as previous one. we can directly write ge q =2 2 Q a Q q (c) The entire system cannot be in equilibrium since both conditions. Solution The force on the point charge Q due to the element dq of the ring ge 1 dqQ dF = along AB 4pe 0 r 2 As for every element of the ring there is symmetrically situated diametrically opposite element.OA2 aP (a 3) a So. T cos q = mg …(i) td While for equilibrium in the plane of equilateral triangle. the components of forces along the axis will add up while those perpendicular a to it will cancel each other. Calculate the time period of oscillation.e..indd 14 3/31/2017 4:31:32 PM . 14  Electrostatics: Part 1 . Show that the motion Le of the negatively charged particle is approximately simple harmonic. tan q = = {as L >> a} 2 L . Hence..(a 3)2 ( 3) L di In g in On substituting the above values of F and tan q in equation (iii) we get: 12 a 3 q2 È 4pe 0 a3 mg ˘ rn = i. F= and tan q = = and as from Figure (C) OA = AD = a sin 60∞ = 4pe 0 a 2 OP 3 3 3 L2 . A particle of mass ‘m’ and having a negative charge ‘Q’ is placed on the axis at a distance of x(x << a) from the centre of the ring. F=. net force on the charge –Q is ng x 1 È Qdq ˘ F = Ú dF cos q = cos q Ú dF. q = Í ˙ ( 3) L mg 4pe 0 a 2 Î 3L ˚ a  Illustration 15    A thin fixed ring of radius ‘a’ has a positive charge ‘q’ uniformly distributed over it. T sin q = 2F cos 30° . F = r Ú 4pe 0 Í.(ii) . Ú dq = 4pe 0 (a2 + x 2 )3 2  …(1) Ce 4pe 0 r 3 {as r = (a 2 + x 2 )1 2 and Ú dq = q} Module-Vol-II_01..2 ˙ Î r ˚ 1 Qx 1 Qqx So. we have 3F  …(iii) tan q = mg vt 1 q2 OA OA 2 2 a Here.L So from equations (i) and (ii). (d) Distance between A and B is reduced. having an insulating handle and initially neutral. Suppose now that a third identical sphere 3. The electrostatic force acting on sphere 2 due to sphere 1 is F .94 × 102 m in 4. At what distance from each other should these charges be placed in oil of relative permittivity 5 to obtain the same force of interaction (a) 8. (c)) and finally removed (Fig. as shown in figure are placed at A and B. (d)). (c) It is moved upwards AB. (b) It is moved towards B.e.4 × 10–2 m (d) 8. the force between them becomes (a) One fourth (b) One half (c) Double (d) Four times In 3. Two small balls having equal positive charge Q (Coulomb) on each are suspended by two insulating strings of equal length L metre. is touched first to sphere 1 (Fig. (b)). Two point charges in air at a distance of 20 cm.kx with k = 4pe 0 a 4pe 0 a 3 . td 1 Qq Qq F= 3 x = . the restoring force will become linear and so the motion is simple harmonic with time period 2p m 4pe 0 ma3 T= = 2p = 2p w k qQ vt Concept Application Exercise 2 aP 1. As the restoring force is not linear. from each other interact with a certain force.94 × 10–2 m (b) 0. Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with  ge their diameters (Fig. The electrostatic force that now acts on sphere 2 has magnitude F’. –q charge is placed at point C midway between A and B..L i. (a)). However. . –q charge will oscillate if (a) It is moved towards A.indd 15 3/31/2017 4:31:32 PM . di 2. Electrostatics: Part 1  15 . if x << a so that x2 << a2. The whole set up is taken in a satellite in to space where there is no gravity (state of weightlessness). What is the ratio F’/F? a ng (a) (b) Ce Module-Vol-II_01. from a hook fixed to a stand. Then the angle (q) between the two strings is rn (a) 0° (b) 90° (c) 180° (d) 0° < q < 180° 5.894 × 10–2 m g (c) 89. The force on any charge is a (a) Zero (b) Kq 2 3 Le a2 Kq 2 Kq 2 (c) (d) 3 3 3a 2 a2 6. Two similar charge of +Q . Three equal charges (q) are placed at corners of an equilateral triangle. the motion will be oscillatory. then to sphere 2 (Fig. When the distance between two charged particle is halved. the spheres repel each other with an electrostatic force of 0.0 m/s2.0 mC are held at separation L = 10. what is the ratio q1/q2? di 9. but the aP net electrostatic force on it from particles 1 and 2 happens to be zero.0 × 10 −7 kg. what was (a) the negative charge on one of them and (b) the positive charge on the other? 11. If the net electrostatic force on particle 3 due to particles 1 and 2 is zero and L23 = 2.0 cm on an x-axis. Particles 1 and 2 are fixed in place.indd 16 3/31/2017 4:31:32 PM . a Le a ge 12. Of the initial charges on the spheres. but particle 3 is free to move. what must be the (a) x and (b) y coordinates of particle 3? Ce Module-Vol-II_01. three charged particles lie on an x-axis. ng If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero.108 N when their centre- g to-centre separation is 50. (a). td .0 cm.0360 N. fixed in place. When the wire is removed. The charges are q1 = q2 = Q and q2 = q1 = q. In the figure. In figure. particles 1 and 2 are fixed in place. Two identical conducting spheres. If the mass of the first particle is 6. what is the ratio q1/q2? In 10.0 m/s2 and that of the second to be 10. what are (a) the mass of the second particle and (b) the magnitude of the charge of each particle? 8. particle 1 of charge +1. In the figure.00L12. Particle 3 is free to move.0 mC and particle 2 of charge −3. In the figure four particles form a square. Two equally charged particles are held 5.L (c) (d) 7. The initial acceleration of the first particle vt is observed to be 6. 16  Electrostatics: Part 1 . The spheres are then connected by a thin conducting wire.0 mm apart and then released from rest. What is Q/q if the net electrostatic force on particles 1 and 4 is zero? (b) Is there any value of q that makes the net electrostatic force on each of the four particles rn zero? Explain. If L23 = L12. attract each other with an electrostatic force of 0. with a positive net in charge. L 14.     E = E + E + E 3 + 1 2 Module-Vol-II_01. Figure (b) gives the x component of that force versus the coordinate x at which particle 3 is placed.50a? . td 13. In Fig. If their net electrostatic force on particle 3 of charge + Q is to be zero. Three particles are fixed on an x axis. What are (a) the sign of charge q1 and (b) the ratio q2/q1? vt aP di (a) (b) 5. Particle 3 (of charge q3 = +8. particle 1 (of charge q1) and particle 2 (of charge q2) are fixed in place on an x-axis. 8. The scale of the x-axis is set by xs = 8. g in a rn Le ‘Electric field’ due to a point charge is the space surrounding it. (a).indd 17 3/31/2017 4:31:33 PM .0 cm. and particle 2 of charge q2 is at x = +a.  Electric Field In ‘The electric field intensity’ (or strength of electric field or electric intensity) at a point in an electric field is the force experienced by a unit positive charge placed at that point. within which electric force can be experienced by another charge. Particle 1 of charge q1 is at x = −a.00 cm apart.net on it. Electrostatics: Part 1  17 . what must be the ratio q1/q2 when particle 3 is at (a) x = +0. a ge ng Superposition of electric field’ (electric field at a point due to various charges): The resultant electric field at any point is equal Ce to the vector sum of electric fields at that point due to various charges.500a and (b) x = +1. provided the presence of this charge does not disturb the field.00 × 10 −19 C) is to be placed on the line between particles 1 and 2 so that they produce a net electrostatic  force F 3. E = F .T rn  Illustration 16    Four point charges all have the same magnitude. is the net electric ng field greater? Statement (i): It is greater in Figure (a).indd 18 3/31/2017 4:31:34 PM . as the figure shows. Ce Module-Vol-II_01. by ‘principle of super-position’ for In discrete distribution:      1 q  E = E1 + E2 +  =  Ei with Ei = r 4pe 0 ri3 i g Electric field intensity is vector quantity. but they do not have the same sign. unit of E is dyne/stat-coulomb. The magnitude of the resultant of two electric fields is given by td E = E12 + E22 + 2 E1E2 cos q .L E2 sin q and the direction is given by tan a = E1 + E2 cos q  ‘Electric field strength or Electric intensity’ ( E ) at a point is the electric force experienced by a unit positive charge at that  vt  point. if either. These charges are fixed to the corners of a rectangle in two different ways. electric field at position r in free space  1 q  1 q di E= r or E = 4pe 0 r 3 4pe 0 r 2 If instead of single charge.r1 ) r2 . 18  Electrostatics: Part 1 . electric field at B due to charge q at A. a Le a ge Consider the net electric field at the centre C of the rectangle in each case. where q0 is positive test charge. SI unit of electric intensity is NC–1 (newton/coulomb)  In CGS system. In vector form. Statement (iii): The field has the same magnitude in both cases. q0  aP q   E = k   3 (r2 .r1  So for a point-charge q. In which of the statements. field is produced by a charge distribution. in  Force MLT -2 MLT -2 Dimensional formula of E : = = = [ MLT -3 A-1 ] Charge ampere × time A. Statement (ii): It is greater in Figure (b). Mathematically. (b). The . However.. 0. In   It leads to a contribution to the net field at C that is shown E 24 as in both parts of the figure. cancel. (b). (a) is less than (not di greater than) the net field in Fig. but combine to produce the field E13 shown in Fig.’ To assess the net field at C.0) kˆ = 3 iˆ + ˆj + kˆ Le with r1 = (32 + 12 + 12 ) = 11 m Æ r2 = (3 . so they contribute individual fields of the same magnitude at C that have opposite directions and. 0).e. and the field created by a negative charge points toward the charge. which is clearly greater than either of these two values alone.0) iˆ + (1 . E = E1 + E2 = ÍÁ + ˜ i + Á + ˜ j+Á + ˜ k˙ 4pe 0 ÍÎË 11 11 3 3¯ Ë 11 11 3 3 ¯ Ë 11 11 3 3 ¯ ˙˚ According to given problem: Ex = 0.0) kˆ = iˆ + ˆj + kˆ ge with r2 = (12 + 12 + 12 ) = 3 m Æ Æ 1 Q ˆ + ˆj + kˆ ] and E = 1 Q So that. E1 = [3 i 2 [ iˆ + ˆj + kˆ ] 4pe 0 (11)3/2 4pe 0 (3)3/2 a Æ Æ Æ 1 ÈÊ 3 ¥ 10 -9 Q ˆ ˆ Ê 10 -9 Q ˆ ˆ Ê 10 -9 Q ˆ ˆ˘ ng And hence. in Fig. Statements (i) and (iii) are incorrect. The field created by a positive charge points away from the charge. (b) but not in Fig.   but in Fig. Each of the individual fields has the same magnitude. If the x-component of the electric field at (3. (b) the charges at corners 1 and 3 are –q and +q. The fact that this contribution to the net field at C is present in Fig. 1. calculate the value of Q. Electrostatics: Part 1  19 . in both cases. respectively. vt note that the charges on corners 2 and 4 are identical in both parts of figure. To see why these answers are incorrect. 1) is zero. which are −q and +q.indd 19 3/31/2017 4:31:35 PM .L directions of the individual fields are different. Is the y-component zero at (3. This is just like the arrangement on corners 1 and 3.0) ˆj + (1 . since the charges all have the same magnitude and are equidistant from C. which was discussed previously. (a) means that the net fields in the two cases are different and that the net field in Fig. aP therefore. Í + ˙=0 4pe 0 ÎÍ 11 11 3 3 ˚˙ Module-Vol-II_01. g  Illustration 17    A charge 10−9 coulomb is located at origin in free space and another charge Q at (2. Solution td The net electric field at C is the vector sum of the individual fields created there by the charges at each corner.0) ˆj + (1 . in Fig. They contribute individual fields of the same magnitude at C that  have the same directions and do not cancel. (a) the charges at corners 1 and 3 are both +q. (a) the net field at C is just E 24 . Moreover. Ce 1 È 3 ¥ 10 -9 Q ˘ i. 1. In Fig. respectively. however. 1)? in Solution  As electric field due to a point charge qi at position ri in vector from is given by rn Æ 1 q Æ Ei = r 4pe 0 ri3 i a Æ here: r1 = (3 .2) iˆ + (1 . ‘Statement (ii) is correct. (b) it is the vector sum of E13 and E 24 . we need to consider the contribution from the charges at corners 2 and 4. But as per Gauss’s law this is not allowed because this closed surface does not enclose any charge. +q). But if the test charge is displaced along OP (a line normal to AB at O). g (b) For the simple configuration of two charges (+q.  [NCERT] di Solution (a) In case.. Consequently. q (charge on the drop) = 12 e = 12(1.26 g/cm3. there should be a restoring force in all directions.indd 20 3/31/2017 4:31:35 PM .81 m/s2. [NCERT] Le Solution Here. then the test charge will experience a restoring force towards the null point when displaced slightly in any direction.81 ¥ 10 -4 mm Ce or r=Í ˙ =Í 2 ˙ Î 4pr g ˚ 3 3 ÎÍ 4 ¥ 3. ˙=. As the drop is held stationary under constant electric field E. The density of the oil is 1. where E = 0) of the configuration. all electric field lines would In be directed inwards towards the null point.6 × 10–19 C) = 19..14 (1. Ey is not zero.. it is necessarily unstable.55 × 104 N/C in a Millikan’s oil drop experiment. force on the drop due to electric a field (qE) = weight of the drop (mg) ng Ê 4p 3 ˆ or qE = Á r rg Ë 3 ˜¯ 1/3 1/3 È 3qE ˘ È 3(1. Estimate the radius of the drop.L 1 È 10 -9 (3 / 11)3/2 ¥ 3 ¥ 10 -9 ˘ 1 2 ¥ 10 -9 Ey = Í .  Illustration 18     aP (a) Consider an arbitrary electrostatic field configuration. there would be a net inward electric flux through a closed surface around the null point. e = 1.e.2 × 10–19 C E = 2.26 g/cm3 = 1.e. the net force F will take it away from the null point along OP. the equilibrium is unstable. Q = -Í ˙ ¥ 3 ¥ 10 -9 coulomb Î 11 ˚ And for this value of Q . If the test charge (q0) is displaced along AB in any direction.55 ¥ 10 4 V/m) ˘ = 0.981 ¥ 10 -6 m = 9.60 × 10–19 C). hence the equilibrium cannot be at all stable.26 × 103 kg/m3 g = 9. In such a situation. A small test charge is placed at a null point (i. (g = 9. rn  Illustration 19    An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.92 ¥ 10 -19 C) (2. 20  Electrostatics: Part 1 . there acts a restoring force on it. the midpoint (O) of the line AB joining the in charges is the null point. But for the stability of the equilibrium.e.81 m/s2 Let r be the radius of the drop.81 m/s ) ˚˙ Module-Vol-II_01. the equilibrium is stable.26 ¥ 10 kg/m ) (9. Show that the equilibrium of the test charge is necessarily unstable. i. π0 4pe 0 ÎÍ 11 11 3 3 ˚˙ 4pe 0 11 11 vt i.55 × 104 V/m ge r (density of oil) = 1. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart. Hence. 3/2 td È3˘ So that. e. i.. Electrostatics: Part 1  21 . density i. (This same technique works for any situation in which charge is distributed along a line or a curve.L field at the given point is the summation of fields of all the elements. with no component perpendicular to that axis (that is. Module-Vol-II_01. we have E x = Ú dE x and E y = Ú dE y . Let us calculate the electric field at a point P that lies on the axis of the ring at a distance x from its centre. So the field at P is described completely by its x-component Ex. Hence that total y-component of field due to this pair of segments is zero. the whole volume of the body. Each segment has charge dQ and ge  acts as a point-charge source of electric field..) The calculation of a  E is greatly simplified because the field point P is one the symmetry axis of the ng ring. we imagine the ring divided into infinitesimal segments of length ds. When we add up the contributions from all such  Ce pairs of segments..indd 21 3/31/2017 4:31:36 PM .e.  Electric Field Due To Continuous Distribution of Charge td To find the field of a continuous charge distribution. charge on For example: Charge on a conducting For example: Non-conducting charged a ring etc.. Field of Ring Charge A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it.  Taking dE = dE x iˆ + dE y ˆj . As shown in the figure. The net   .e. Let dE be the electric field from one  such segment. we divide the charge into infinitesimal charge elements. no y-component or z-component). E = Ú dE Continuous charge distribution Linear charge distribution Surface charge distribution Volume charge distribution vt In this the distribution charge is distributed In this the distribution charge is distributed In this the distribution charge is distributed in on a line. the net electric field at P is then the sum of all contributions dE from all the segments that make up the ring. Relevant parameter is l which is sphere. If we consider two ring segments at the top and bottom of the ring. E = Ú dE . on the surface. aP For example: Charge on a wire. charge charge charge l= s= r= volume di length area Q Q Q s= r= l= 4 2p R 4p R 2 p R3 In 3 g Spherical shell Circular charged ring Non conducting sphere in dq Each infinitesimal charge element is then considered as a point charge and its field is given by: dE = . as a point charge and electric field dE is determined due to this charge at given point. we see that  the contributions dE to the field at P from these segments have the same x-component but opposite y-components. Relevant sphere. a Le 7. 6. Each infinitesimal  charge element is then considered. parameter is s which is called surface charge volume charge density i. charge on a sheet etc. the total field E will have only a component along the ring’s symmetry axis (the x-axis). 4pe 0r 2 rn   At a point distant r from the element. Relevant parameter is r which is called called linear charge density i.e. the net field is the summation of fields of all the elements. Hence the magnitude of td 1 dQ this segment’s contribution to the electric field at P is dE = 4pe 0 x + a 2 2 x x . hence it should have certain maximum value between x = 0 and x = • (or x = –• ) g ∑ If we maximize the equation (i) we can get the value of xm as well as Emax. To calculate Ex note that the square of the distance r from a ring segment to the point P is r2 = x2 + a2. all the factors on the right side except dQ are constant and can be taken outside the integral. ∑ For x = 0. this conclusion may be arrived at the symmetry consideration. in a rn Le ge d ÏÔ 1 x Ô¸ For maximum value of Ex. The integral of dQ is just the total charge Q is just the total charge Q.. Ì Q ˝=0 dx ÓÔ 4pe 0 ( x 2 + a 2 )3/2 ˛Ô 3 ( x 2 + a 2 )3/2 ◊1 . from the ring the electric field will be zero. 22  Electrostatics: Part 1 .3x 2 = 0 fi x = ± 2 Ce 1 Ê 2Q ˆ and the maximum value of the electric field is Ea(max) = 4pe 0 ÁË 3 3 R 2 ˜¯ Module-Vol-II_01. ∑ At a large distance. we integrate this expression over all segments of the ring: aP 1 xdQ Ex = Ú 4pe 0 ( x + a 2 )3/2 2 Since x does not vary as we move from point to point around the ring. and we finally get di Æ 1 xQ E = E x iˆ = iˆ  …(i) 4pe 0 ( x + a 2 )3/2 2 In ∑ Electric field is directed away from positively charged ring. E = 0.indd 22 3/31/2017 4:31:37 PM .L Using cos a = = 2 . the component dEx of this field along the x-axis is r ( x + a 2 )1 2 1 dQ x 1 xdQ dE x = dE cos a = = 4pe 0 x + a 2 2 2 x +a 2 4pe 0 ( x + a 2 )3/2 2 vt To find the total x-component Ex of the field at P.x ◊ ( x 2 + a 2 )1/2 ◊ 2 x a 2 =0 ( x 2 + a 2 )3 ng a ( x 2 + a2 ) . We will integrate dEx as well as dEy in limits q = –q1 to q = +q2. Let us find the electric field at point D on the x-axis at a distance r0 ge from the origin. we obtain di + p /2 lR l E= Ú 4pe 0 R 2 cos q dq = 2pe 0 R . q1 = q2.cos q 2 ) -q1 l sin q a For a symmetrical arc. let the length of a typical segment at height l be dl. that subtends an angle dq at the centre of the ring.r. aP dQ l ( R dq ) cos q dE x = dE cos q = 2 cos q = 4pe 0 R 4pe 0 R 2 On integrating the expression for dEx. lying along the y-axis. vt For each differential element in the upper half of the ring. and x-components remain. each of which acts as a point charge. Field of Line Charge Positive electric charge Q is distributed uniformly along a line.indd 23 3/31/2017 4:31:37 PM . The y-components of field due to these symmetric elements cancel out. angle q.  Illustration 20    A uniformly charged wire. Electrostatics: Part 1  23 . w. We divided the line charge into infinitesimal segments. If the charge is distributed uniformly with the linear charge density l. there corresponds a symmetrically placed charge element in the lower half plane. l = Q pR fi E= Q 2p e 0 R 2 2 In If we consider the wire in the form of an arc as shown in the figure.t. in limits q = –p/2 to q = +p/2. At point D the differential electric field dE created by this element. This element creates a field dE which makes an angle q at the centre as shown in figure. if q1 and q2 are different. Hence the charge dQ in a segment of length a dl is dQ = l dl.L We consider a differential element dl on the ring. Thus Ey vanishes and E x = 2pe 0 R Le 8. the symmetry consideration is not useful in cancelling out g x and y components of the fields. Find the td electric field generated by the semicircle at the centre? Solution . ng dQ l dl dE = 2 =  …(i) 4pe 0r 4pe 0r 2 In triangle AOD. OA = OD tan q Ce l = r0 tan q. +q 2 in lR l Ex = Ú 4pe 0 R 2 cos q dq = 4pe 0 R (sin q1 + sin q 2 ) -q1 +q 2 rn lR l Ey = Ú 4pe 0 R 2 sin q dq = 4pe 0 R (cos q1 . is laid in the form of a semicircle of radius R. dl = r dq.p /2 In terms of charge. linear charge density l. Differentiating this equation with respect to q dl = r0 sec2 q dq Module-Vol-II_01. for a very long wire (infinitely long wire). where | Ey | tan q = = 1. l dq td Substituting the value of dl in equation (i).p /2          Using a symmetry argument. dE = 4pe 0r0 l cos q dq l sin q dq Field dE has components dEx. q = 45∞ | Ex | a  Illustration 21    What is the electric field at any point on the axis of a charged rod of length L and linear charge density l'? ng The point is separated from the nearer end by a.cos q 2 ) g -q1 in ∑ “If we wish to determine field at the end of a long wire. if we place a positive test charge at D. 24  Electrostatics: Part 1 .” rn l È Ê p ˆ˘ l Ex = Ísin (0) + sin ÁË ˜¯ ˙ = 4pe 0 r0 Î 2 ˚ 4pe 0 r0 l È Ê p ˆ˘ l and E y = Ícos(0) + cos ÁË ˜¯ ˙ = a 4pe 0 r0 Î 2 ˚ 4pe 0 r0 Le Æ Magnitude of resultant field E : Æ 2l | E | = E x2 + E y2 = 4pe 0r0 ge Æ E makes an angle q with the x-axis. the upper half of the line of charge pushes downward on it.p /2 . ∑ “If the wire has finite length and the angle subtended by ends of wire at a point are di q1 and q2. Note that as the length of wire 2 2 increases. the limits of integration would change. vt + p /2 + p /2 l cos q dq l l sin q dq l Ex = Ú 4pe 0r0 = 2pe 0r0 and E y = Ú 4pe 0r0 =0 Thus E = E x = 2pe 0r0 aP .indd 24 3/31/2017 4:31:38 PM .” +q 2 l cos q dq l Ex = Ú = (sin q1 + sin q 2 ) In -q1 4pe 0r0 4pe 0r0 +q 2 l sin q dq l Ey = Ú 4pe 0r0 = 4pe 0r0 (cos q1 . the angle q increases. Ce Module-Vol-II_01. we may substitute q1 = 0 and q2 = p /2 in the expressions for Ex and Ey. to q = + we obtain Ex and Ey. we could have guessed that Ey would be zero. it approaches p /2. and the lower half pushes up with equal magnitude. dEy given by dE x = and dE y = .L 4pe 0 r0 4pe 0 r0 p p On integrating expression for dEx and dEy in limits q = . x from the point. ˙ = Í 1. or dA = 2pr dr. simultaneously adding charge so that the surface charge density s (charge per unit area) is constant. z = x2 + r2. so the charge of ring is dQ = s (2p r dr). In this figure. we must integrate from 0 to R (not from –R to R): rn R 1 (2ps r dr ) x Ex = Ú dE x = Ú dE x = 4pe 0 ( x 2 + r 2 ) 3/2 0 a Remember that x is a constant during the integration and that the integration variable is r. Electrostatics: Part 1  25 .˙ Î x ˚a = kl Í Î a + L + ˙ a ˚ . the charge is assumed to be positive. The dQ In charge per unit area is s = . the electric field due to the ring has no components perpendicular to the axis. In the limit that R is much larger than the distance x of the field point from the disk. where we seek to find the electric field. To include the whole disk. and the total field has Ey = Ez = 0. The field component dEx at point P due to charge dQ of a single of radius r g 1 (2ps r dr ) x dE x = in 4pe 0 ( x 2 + r 2 )3/2 To find the total field due to all the rings. then E = s From (i) E x = Í 1. dq = ldx a+L a+L Then l dx 1 È 1˘ È -1 1˘ dE = k ◊ x2 or E = k l Ú x2 dx = k l Í . the result is Le sx È 1 1˘ s È x ˘ Ex = Í– + ˙= Í1 – ˙ (i) 2e 0 ÍÎ x 2 + R 2 x ˙˚ 2e 0 ÍÎ ( x 2 + R 2 ) ˙˚ Finding the electric field on the axis of a uniformly charged disk. We can represent this charge distribution as a collection of concentric rings of charge. ˙ x 2e 0 R ˙ 2e 0 Í 2 2 R ˙ 2e 0 Ce Í x 1+ 2 1+ 2 ÎÍ x ˚˙ ÎÍ x ˚˙ Module-Vol-II_01. We already know how to find the field of a single ring on its axis of symmetry. We’ll let you work out the details. so all we have to do is add di the contribution of all the rings. Hence at point P in the figure. and outer radius r + dr. The situation is shown in figure. dx at a distance. inner radius r. A point on ge the symmetry axis of a uniformly charged ring. Solution td Consider an element.L a l È1 1 ˘ È 1 ˘ Thus E= Ía . Its area dA is approximately equal to its width dr times its circumference 2pr. at a point along the axis of the disk a distance x from its centre. The elemental charge. dEy = dEz = 0 for each ring. Suppose we keep on increasing the radius R of the a disk. The integral can be evaluated by use of the substitution. The situation becomes the electric field near infinite plane sheet ng of charge. a typical ring has charge dQ. dA or dQ = 2ps r dr. Field of Uniformly Charged Disk aP Let us find the electric field caused by a disk of radius R with a uniform positive surface charge density (charge per unit area) s.          As R > > x.L + a˙ Ík = ˙ 4p Œo Î ˚ Î 4pe 0 ˚ vt 9. we integrate dEx over r. P. È ˘ È ˘ Í ˙ Í ˙ s Í x ˙ s Í 1 ˙ . Again we can ask what happens if the charge distribution gets very large.indd 25 3/31/2017 4:31:39 PM . (R > > x). As shown in figure. 26  Electrostatics: Part 1 . Our final result does not contain the distance x from the plane. This is correct but rather surprising result this means. td • That the electric field produced by an infinite plane sheet of charge is independent of the distance from the sheet. • Thus the field is uniform; its direction is everywhere perpendicular to the sheet and away from it. .L • Infinite plane sheet of charge is a hypothetical case in real practice there is no such infinite plane sheet of charge. Again, there is no such thing as an infinite sheet of charge, but if the dimensions of the sheet are much larger than the distance x of the observation point P from the sheet, the field is very nearly the same as for an infinite sheet. vt 10. Field of Two Oppositely Charged Sheets Two infinite plane sheets are placed parallel to each other, separated by a distance d (as shown in figure). The lower sheet has a uniform positive surface charge density s, and the upper sheet has a uniform negative surface charge density –s with the same aP magnitude. Let us find the electric field between the two sheets, above the upper sheet, and below the lower sheet. di In The situation described in this example is an idealization of two finite, oppositely charged sheets, like the plates shown in the g figures. If the dimensions of the sheets are large in comparison to the separation d, then we can do good approximation, consider the sheets to be infinite in extent. We know the field due to a single infinite plane sheet of charge. We can then find the total field by in using the principle of superposition of electric fields. Let sheet 1 be the lower sheet of positive charge, and let sheet 2 be the upper   sheet of negative charge; the fields due to each sheet are E1 and E2 , respectively and both have the same magnitude at all points, s rn no matter how far from either sheet E1 = E2 = . 2e 0   At all points, the direction of E1 is away from the positive charge of sheet 1, and the direction of E2 is towards the negative a charge of sheet 2. These fields, as well as the x-axis and y-axis, are shown in figure. At points between the sheets, the fields reinforce each other; at points above the upper sheet or below the lower sheet, the fields cancel each other. Thus the total field is Le Ï0 above the upper sheet Æ Æ Æ Ô Ôs ˆ E = E1 + E2 = Ì j between the sheets Ô e0 ÔÓ 0 below the lower sheet ge Because we considered the sheets to be infinite, our result does not depend on the separation of d. a ng Ce Module-Vol-II_01.indd 26 3/31/2017 4:31:39 PM Electrostatics: Part 1  27 . Symmetry play very important role in problem solving. Electric field is in the direction along the line which divides the td change distribution symmetrically. .L vt aP Enet = Ú dE cos q Enet = Ú dE cos q di In g in Enet = Ú dE cos q rn Enet = Ú dE cos q Two point charges Three point charges at the corner of an equilateral triangle a Le ge Æ Æ | E1 | = | E2 | Here, a Enet = 2 | E1 | cos q ng Here electric field at P due to charges (1), (2) and (3) are equal. Æ Æ Æ | E1 |= | E2 | = | E3 | Æ Ce Enet = 3 | E1 | cos q Hence, Module-Vol-II_01.indd 27 3/31/2017 4:31:40 PM 28  Electrostatics: Part 1 . Some useful results: td A charged rod having charge density l of fixed length Semi-infinite rod having charge density l .L vt aP l Ex = (sin q1 +sin q 2 ) 4pe 0r0 l di Ey = (cos q1 - cos q 2 ) 4pe 0r0 Semicircular ring having charge density l Quarter circular ring having charge density l In g in rn l Ex = 4pe 0 r a l l Ex = Ey = 2pe 0r 4pe 0r Le Ey = 0 Infinite line charge Charged ring a ge ng 1 xQ Ex = 4pe 0 ( x + a 2 )3/2 Ce 2 l Ex = 2pe 0r Ey = 0 Ey = 0 Module-Vol-II_01.indd 28 3/31/2017 4:31:41 PM  Motion of Charged Particle in an Electric Field di When charged particle initially at rest is placed in the uniform field: Let a charge particle of mass m and charge Q be In initially at rest in an electric field of strength E g in rn (i) Force and acceleration: The force experienced by the charged particle is F = QE. (A)]. v 2 = 0 + 2 ¥ ¥ s v2 = Ì∵ E = ˝ fi v = m m Ó s˛ m a QE QEt 2QV (iii) Momentum: Momentum p = mv. (B)] V = Potential difference between A and B. S = Separation between A and B E QEt (a) By using v = u + at . v = 0 + Q t. fi v = m m ge QE 2QV Ï V¸ 2QV (b) By using v 2 = u 2 + 2as. p = mv. F QE a Acceleration produced by this force is a = = m m Since the field E is constant the acceleration is constant.indd 29 3/31/2017 4:31:42 PM . Le (ii) Velocity: Suppose at point A particle is at rest and in time t. Ey = 0 aP 2e 0 Í s Î ( x 2 + R 2 ) ˙˚ Ex = 2e 0 fi Ey = 0 11. Charged disc Infinite sheet of charge td .L vt s È x ˘ Ex = Í1 – ˙ . p = m ¥ = or p = m ¥ = 2 mQV m m m ng 1 2 1 (QEt )2 Q 2 E 2 t 2 (iv) Kinetic energy: Kinetic energy gained by the particle in time t is K = mv = m = 2 2 m 2m or 1 2QV K= m¥ = QV Ce 2 m When a charged particle enters with an initial velocity at right angle to the uniform field: When charged particle enters perpendicularly in an electric field. it describes a parabolic path as shown. thus motion of the particle is uniformly accelerated. Electrostatics: Part 1  29 . Positive charge experiences force in the direction of electric field while negative charge experiences force in the direction opposite to the field [Fig. it reaches the point B [Fig. Module-Vol-II_01. Since the motion of the particle is accelerated along y-axis. these are negatively charged.e. vx = u and vy = m di  Q2 E 2t 2 So v = | v | = vx2 + vy2 = u 2 + m2 If b is the angle made by v with x-axis then tan b = vy vx = In QEt mu . Ce Module-Vol-II_01.L 1 S = ut + at 2 2 1 2 We have u = 0 (along y-axis) so y = at vt 2 i. (i) Equation of trajectory: Throughout the motion particle has uniform velocity along x-axis and horizontal displacement (x) is td given by the equation x = ut . Ee Ê eˆ a= =E Á ˜ m Ë m¯ ge Since particle (3) is deflected the most. The plates are sufficiently long ng and have separation 2 cm. a  Illustration 23    A particle having charge that on an electron and mass 1. Find the maximum value of velocity of particle not to hit the upper plate.indd 30 3/31/2017 4:31:42 PM . e/m (charge to mass ratio) is the highest for particle (3). g  Illustration 22    Figure shows tracks of three charged particles in a uniform electrostatic field. we will use equation of motion for uniform acceleration to determine displacement y. displacement along y-axis will increase rapidly with time (since y µ t2) x From displacement along x-axis t = aP u 2 1 Ê QE ˆ Ê x ˆ 2 So y = Á ˜ Á ˜ . 30  Electrostatics: Part 1 . which is attracted towards the negative plate is positively charged. as E is same for all the particles. Further. Take electric field between the plates = 103 V/m directed upward.6 × 10–30 kg is projected with an initial speed u at an angle = 45° to the horizontal from the lower plate of a parallel plate capacitor as shown in figure. From . this is the equation of parabola which shows y µ x 2 Ë m ¯ Ë u¯ QEt (ii) Velocity at any instant: At any instant t.. Give the sign of the three charges. it experiences the highest acceleration (a). Particle (3). If a is the acceleration of a charged particle in the electric field. Which particle has the highest charge to mass ratio? [NCERT] in a rn Solution Le Since particles (1) and (2) are attracted towards the positive plate. 0 ˆj hence |a| = . 2 ¥ 1. Electrostatics: Part 1  31 . Solution td Resolving the velocity of particle parallel and perpendicular to the plate.. The particle accelerates towards the other plate a distance d away. The uniform field is created between two parallel plates as shown in figure.qE0 ˆj F ^ d E0 i The force is constant.  Electric Dipole An electric dipole is a system of two equal and opposite point charges separated by a very small and finite distance. Solution Æ In q g The applied electric field is E = .indd 31 3/31/2017 4:31:43 PM . Le Using equation.6 ¥ 10 -19 ¥ 103 ¥ 2 ¥ 10 -2 u^ = 2ay = = 2 ¥ 106 m/s fi u = 2u^ = 2 2 ¥ 106 m/s 1. with y £ d where d. F = qE = . aP \  Maximum velocity for the particle not to hit the upper plate. and so the acceleration is constant as well rn Æ Æ F qE qE0 X a = = . (i) Dipole axis: Line joining negative charge to positive charge of a dipole is called its axis. u u u = u cos 45∞ = and u^ = u sin 45∞ = . distance between the plates. in downward direction – m m m Due to constant acceleration the particle moves in y-direction. a = . where m. i. Ce (ii) Equatorial axis: Perpendicular bisector of the dipole is called its equatorial or transverse axis as it is perpendicular to length.L 2 2 Force on the charged particle in downward direction normal to the plate = eE eE \  Acceleration. Module-Vol-II_01. Determine the speed at which it strikes the opposite plate. s = d and a =  …(ii) m Ê qE ˆ ge v2 = 0 + 2 Á 0 ˜ d Ë m ¯ fi 2qE0 d v= a m ng 12. mass of charged particle vt m The particle will not hit the upper plate.E0 ˆj + Æ Æ ^ j in The force experienced by the charge q. if the velocity component normal to plate becomes zero before reaching it. v2 = u2 + 2as …(i) qE 0 Here u = 0. the problem is analogous to motion of a mass released from rest a in a gravitational field. It may also be termed as its longitudinal axis.e.6 ¥ 10 -30 di  Illustration 24    A particle of mass m and charge q is released at rest in a uniform field of magnitude E. 0 = u2^ – 2ay. 32  Electrostatics: Part 1 .1  Electric Field Intensity due to an Electric Dipole at a Point on the Axial Line rn A line passing through the negative and positive charges of the electric dipole is called the axial line of the electric dipole. It is denoted as p and is defined as the product of the magnitude of either of the charge and the dipole length. td (iv) Dipole moment: It is a quantity which gives information about the strength of dipole. p = q (2l ) Its SI unit is coulomb-metre or Debye (1 Debye = 3. Since E1 > E2.  Electric Field Due to a Dipole 13. HCl. Thus. (iii) Dipole length: The distance between two charges is known as dipole length (L = 2l). Let P be an observation point on the axial line such a that its distance from the mid-point O of the electric dipole is r.3 × 10–30 C × m) and its dimensions are M0L1T1A1.e.     di ∑ Water (H2O). We are interested to calculate the intensity of electric field at P. Chloroform (CHCl3). CO molecules are some example of permanent electric dipole.L   i.E2 = ◊ . ◊ 4pe 0 K (r . Suppose an electric dipole AB is located in a medium of dielectric constant K (as shown in figure). . It is a vector quantity and is directed  from negative charge to positive charge along the axis. aP ∑ When a dielectric is placed in an electric field.l ) 2 4pe 0 K (r + l )2 Module-Vol-II_01. vt Note: ∑ A region surrounding a stationary electric dipole has electric field only.l )2 a 1 q and E2 = ◊ {along the direction OB} ng 4pe 0 K (r + l )2 The intensities E1 and E2 are along the same line but in opposite directions. Le ge 1 q E1 = ◊ {along the direction OX} 4pe 0 K (r . Ce 1 q 1 q E = E1 . In g in 13. hence resultant intensity E at the  point P will be equal to their differences and in the direction AP. Ammonia (NH3).indd 32 3/31/2017 4:31:44 PM . its atoms or molecules are considered as tiny dipoles. Let the dipole consists of two point charges of -q and +q coulomb separated by a short distance 2l metre. Electrostatics: Part 1  33 . ge OA OA l cos q = = = PA (OP 2 + OA2 )1/2 (r 2 + l 2 )1/2 1 q 2l 1 2ql a \ E= ◊ ◊ = ◊ 4pe 0 K (r 2 + l 2 ) (r 2 + l 2 )1/2 4pe 0 K (r 2 + l 2 )3/2 ng But 2ql = p = electric dipole moment 1 p \ E= ◊ 4pe 0 K (r 2 + l 2 )3/2 Ce If l is very small as compared to r(l << r). In 14. the direction of electric field E is in the direction of p i. then l2 can be neglected in comparison to r2.l ) ˚ 4pe 0 K Î (r . Let us assume again that the medium between the electric dipole and the observation in point has dielectric constant K.l ) ˚ 1 2 pr .. then l2 can be neglected in comparison to r2. 1 2p fi E= ◊ 4pe 0 r 3  i.e.l ) If l is very small compared to r(l << r).L But 2ql = p = electric dipole moment. Then the electric field intensity at the point P due to a short dipole is given by vt 1 2 pr 1 2p E= ◊ 4 = ◊ 3 4pe 0 K r 4pe 0 K r aP In vacuum.  Electric Field Intensity Due to an Electric Dipole at a Point on the Equatorial Line An equatorial line of the electric dipole is a line perpendicular to the axial line and passing through a point mid-way between charges. È 4lr ˘ È 2 (2ql ) r ˘ td q 1 E= Í 2 2 2˙= Í 2 21 2 ˙ 4pe 0 K Î (r . Le E = E1 cos q + E2 cos q [sine components cancel out] 1 q 1 q 1 q E= ◊ 2 2 cos q + ◊ 2 2 cos q = ◊ 2 2 ◊ 2 cos q 4pe 0 K (r + l ) 4pe 0 K (r + l ) 4pe 0 K (r + l ) But from the figure. 1 q E1 = ◊ 2 2 {along the direction PD} 4pe 0 K (r +l ) rn 1 q and E2 = ◊ {along the direction PC} 4pe 0 K (r 2 + l 2 ) a The magnitude of E1 and E2 are equal but directions are different.e.indd 33 3/31/2017 4:31:45 PM . g Let us now suppose that the observation point P is situated on the equatorial line of dipole such that its distance from mid-point O of the electric dipole is r (as shown in figure). parallel to the axis of dipole from the negative charge towards the di positive charge.. Then the electric field intensity at the point P due to a short dipole is given by Module-Vol-II_01. fi E = ◊ 2 2 2 4pe 0 K (r . the direction of electric field E is opposite to the direction of p i.t.  Electric Field Intensity Due to a Short Distance Dipole at Some General Point  Let AB be a short electric dipole of dipole moment p (directed from B to A). 15. then K = 1 and E = ◊ 3 ..r. hence in vector form we can write   1 p 1 p E= ◊ 3 ( -iˆ ) = .indd 34 3/31/2017 4:31:46 PM .   We know that dipole moment of a dipole is a vector quantity. 2 2 Ê 2 p cos q ˆ Ê p sin q ˆ p p ge or. 1 p 1 p td E= ◊ = ◊ 4pe 0 K (r 2 )3/2 4pe 0 K r 3 1 p If dipole is placed in air or vacuum.. The distance of observation point P w. P lies on the equatorial line of dipole with moment p2 . then a E2 ( p sin q / 4pe 0r 3 ) 1 tan f = = = tan q ng E1 (2 p cos q / 4pe 0r 3 ) 2 16.r. anti-parallel to the axis of dipole from the positive charge aP towards the negative charge. It can be resolved into two rectangular components p1 and p2 as    In shown in figure. Dipole in a Uniform Electric Field Ce 16.   The magnitude of p1 and p2 are p1 = p cos q and p2 = p sin q. 34  Electrostatics: Part 1 .1 Torque When a dipole is placed in a uniform field as shown in figure the net force on it Module-Vol-II_01. .t.L 4pe 0 r As direction of resultant electric field is along the negative x-axis.e. so that p = p1 + p2 .  It is clear from figure that point P lies on the axial line of dipole with moment p1. ◊ 3 vt 4pe 0 r 4pe 0 r  i. Hence. Hence magnitude g   of the electric field intensity E1 at P due to p1 is in 1 2 p cos q E1 = ◊ {along OC} (i) 4pe 0 r3  rn Similarly. di axis of dipole is q.e. mid-point O of the dipole is r and the angle made by the line OP w. We are interested to find the electric field at some general point P. magnitude of electric field   intensity E2 at P due to p2 is 1 p sin q E2 = ◊ (ii) a {opposite to} 4pe 0 r3    Le Hence resultant intensity at P is E = E1 + E2    Magnitude of E is: E = ( E12 + E22 ) (as E1 and E2 are mutually perpendicular). E= Á 3 ˜ +Á 3˜ = 3 4 cos 2 q + sin 2 q or E = 1 + 3cos 2 q Ë 4pe 0 r ¯ Ë 4pe 0 r ¯ 4pe 0 r 4pe 0 r 3   If the resultant field intensity vector E makes an angle with the direction of E1. 0 cm. there will be an SHM Ce Here. are connected by a massless rod of  ge o  electric field at an angle q with the E (q ≈ 0 ).w 2q Ë I ˜¯ As torque is proportional to 'q' and oppositely directed. p = q. Solution a t = pE sin q (as q → 0.   Illustration 25    Find the force on a small electric dipole of dipole moment p due to a point charge Q placed at a distance r.   FR = ÈÎqE + ( .. torque is clockwise) ng Ê pE ˆ fi a = -Á q = 0 . L. value of t will be when sin q = 1 i. aP Solution Electric field of a point charge is a non-uniform electric field. What maximum torque does the field exert on the dipole? Solution a t = pE sin q = q ¥ 2a ¥ E sin q Max. The dipole is placed rn in an external field of 105 N/C. Le \ t max = 10 -7 ¥ 2 ¥ 10 -2 ¥ 10 5 ¥ 1 = 2 ¥ 10 -4 N-m  Illustration 27    Two tiny spheres. t = pE sin q {as p = q 2l}    . sin q → q) fi t = –(pE)q (If we assume angular displacement to be anti-clockwise. They are placed in a uniform system aligns itself parallel to the E .1 mC separated by a distance of 2.e.q) E ˘˚ = 0 td While the torque t = qE ¥ 2l sin q i. The electric field of small dipole at a distance r is E = . each of mass M.indd 35 3/31/2017 4:31:47 PM . Electric field at a distance x from the point charge is 1 Q dE 1 2Q E= fi =- 4pe 0 x 2 dx 4pe 0 x 3 di Magnitude of force on the dipole dE 1 2 pQ In F= p = dx x =r 4pe 0 r 3 Same can be calculated as force on the point charge due to dipole which is same as the force on dipole due to point charge 1 2p g (Newton’s 3rd law). I = M (L/2)2 + M (L/2)2 = ML2/2 Module-Vol-II_01. in 4pe 0 r 3  Illustration 26    An electric dipole consists of two charges of 0.e. 1 2 pQ 4pe 0 r 3 Hence force on the point charge Q is F = .L or t = p¥E From the expression it is clear that couple acting on a dipole is maximum (= pE) when dipole is perpendicular to the field and minimum (= 0) when dipole is parallel or antiparallel vt to the field. Electrostatics: Part 1  35 . and charges +q and –q respectively.L and moment of inertia. Calculate the minimum time in which the length. The common centre a of rings lies at origin and the common axis coincides with z-axis. 36  Electrostatics: Part 1 . cross-sectional area A and Young’s modulus Y. one of radius R and total charge +Q and the second of radius 2R and total charge .qE = q DE È DE ˘ dE È DE dE ˘ dE = qÍ 2a ˙ = 2aq Í as = ˙ =p   Î Dx ˚ dx Î Dx dx ˚ dx Æ di Æ dE |F | = p dx where dE dx is the gradient of the field in the x-direction.. The charge is Le uniformly distributed on both rings. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms–1.e. There is a uniform horizontal electric field of 104 N/C. lie in x-y plane (i. Neglecting gravity. Two concentric rings. the magnitude of extension in this rod is Ce Module-Vol-II_01. in vector from is equal to: (a) 90( -3 iˆ + 4 ˆj ) V/ m (b) 90(3 iˆ .2  Net Force on a Dipole in a Non-Uniform Field Æ Æ Suppose an electric dipole with dipole moment P is placed in a non-uniform electric field E = Eiˆ that points along x-axis. td I \ Time period T = 2p pE T . A point charge 50 µC is located in the x-y plane at the position vector r0 = (2 iˆ + 3 ˆj ) m. The electric field at the point of in  position vector r = (8 iˆ .4 ˆj ) V/ m rn (c) 900 ( -3 iˆ + 4 ˆj ) V/ m (d) 900(3 iˆ . The electric field at the position of negative charge is E and at the position of positive charge ( E + DE ). R (a) R (b) 2 2 ge R (c) (d) 2R 2 2 3. z = 0 plane). The rod is placed in space having uniform electric field of magnitude E and directed parallel to length of the rod as shown. Net force acting on the dipole is then aP F = q ( E + DE ) .4 ˆj ) V/ m 2. A light weight particle of charge Q is fixed at one end of an electrically insulated uniform elastic rod of natural length L.indd 36 3/31/2017 4:31:48 PM .L The minimum time required to align itself is 4 16. In Concept Application Exercise 3 g  1. At what distance from origin is the net electric field on z-axis zero. Let vt E depend only on x.5 ˆj ) m.8Q. a (a) the horizontal range of the particle is 10 m (b)  the time of flight of the particle is 2 s ng (c) the maximum height reached is 5 m (d)  the horizontal range of the particle is 0 4. An electric dipole in a uniform electric field experiences (a) Force and torque both (b) Force but no torque (c) Torque but no force (d) No force and no torque Ce 10.indd 37 3/31/2017 4:31:49 PM . A point P is at a distance of 20 cm from this origin such that p OP makes an angle with the x-axis. If EA. as shown. An electric dipole is placed along the x-axis at the origin O. Three points A.L 4YA 5. the value of q would be 3 ge p Ê 3ˆ (a) p (b) + tan -1 Á ˜ 3 3 Ë 2 ¯ 2p Ê 3ˆ (d) tan -1 Á a (c) ˜ 3 Ë 2 ¯ ng 9. td QEL QEL (a) (b) YA 2YA (c) QEL (d) None of these . If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal. B and C respectively then aP (a) EA > EB > EC (b) EC > EB > EA (c) EB > EA = EC (d) EB > EA > Ec 6. Electrostatics: Part 1  37 . B and C of an equilateral triangle of side a as shown in the adjoining figure. and 2 2 2 vt respectively from their common centre. then x : y is di (a) 1 : 1 (b) 1: 2 (c) 1 : 2 (d) 3 2 :1 In 7. EB and EC are magnitudes of the electric fields at points A. If the electric field at P makes an angle q with x-axis. Consider two thin uniformly charged concentric shells of radii r and 2r having charges Q r 3r 5 and –Q respectively. (– q) and (– q) are placed at the corners A. A point charge placed at any point on the axis of an electric dipole at some large distance experiences a force F. The force acting on the point charge when its distance from the dipole is doubled is Module-Vol-II_01. B and C are marked at distances . Three charges of (+2q). Then the dipole moment of this combination is g in a rn (a) qa (b) Zero 2 Le (c) q a 3 (d) qa 3 8. 15. 3). Module-Vol-II_01.L 4 8 11. A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. –1) and the other shell S2 has radius 6 m and charge 3 mC and centre at (–2. At a point P. What will be minimum time. in terms of l 2 Qb a and X1. Find the electric field (in vector form) at point (1. Two non-conducting spherical shells are uniformly charged. The field strength to the right of the charge Qb ge on the line that passes through the two charges varies according to a law that is represented graphically in the figure. There is a non-uniform electric field E = a ( x + l )iˆ l Ce where a is a constant. far away from the dipole. 1. –1. Two point charges are placed at point a and b. 38  Electrostatics: Part 1 . Another point particle of the same mass is attached to other end of the rod. 2. Find the resultant electric force in mN on the loop if l = 10 cm. One shell S1 having radius 5 m and charge –2mC has centre at (–1. a = 2 N/C and charge density l = 2 mC/m. A square loop of side ‘l’ having uniform linear charge density ‘l’ is placed in ‘xy’  plane as shown in the figure. OP makes an angle q with the x-axis then In g in rn (a) tan q = 3 (b) tan q = 2 a 1 (c) q = 45o (d) tan q = 2 Le 13. The two particles carry charges +q and –q respectively. the electric field is parallel to y-axis. An electric dipole is placed at the origin O and is directed along the x-axis. needed for the rod to become parallel to the field after it is set free? aP mL p mL (a) t = 2p (b) t = 2 pE 2 2qE 3p mL 2mL (c) t = (d) t = p di 2 2 pE qE 12. –1). Find the signs of the charges and ratio of magnitudes of charges Qa and the distance X of the point from b where the field is maximum.indd 38 3/31/2017 4:31:50 PM . 14. This arrangement is held in a region of a uniform vt electric field E such that the rod makes a small angle q (say of about 5 degrees) with the field direction (see figure). td F (a) F (b) 2 F (c) F (d) . The electric field is taken positive if its direction is towards right and negative if its ng direction is towards left. L the properties of electric lines of force. Lines of Force td It has been found quite convenient to visualize the electric field in terms of lines of force. vt (ii) Tangent to the field line at any point gives the direction of the field at that point. di (v) Field lines do not exist inside a conductor. g in rn (vii) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. Electrostatics: Part 1  39 .indd 39 3/31/2017 4:31:51 PM . also the density of field lines is proportional to the strength of the electric field. also number of electric a lines at force linked with QA are more than those linked with QB hence |QA| > |QB|. (ix) If the lines of forces are equidistant and parallel straight lines the field is uniform and if either lines of force are not equidistant or straight line or both the field will be non-uniform. For example. Followings are . f = EA). The line of force in an electric field is a curve such that the tangent at any point on it gives the direction of the resultant electric field strength at that point. (viii) Number of lines of force per unit area normal to the area at a point represents magnitude Le of intensity (concept of electric flux i. aP (iv) Field lines are always normal to conducting surface. In (vi) The electric field lines never form closed loops (while magnetic lines of forces form closed loop). (iii) Field lines never cross each other. a ge ng Ce Module-Vol-II_01. see the following figures. In the following figure electric lines of force are originating from A and terminating at B hence QA is positive while QB is negative. 17.1  Properties of Electric Lines of Force (i) Electric field lines come out of positive charge and go into the negative charge..e. 17. Their working suffers and they may start misbehaving under the effect of unwanted fields. (iii) An earthed conductor also acts as a screen against the electric field. induced positive charge neutralizes and the field in the region beyond AB disappears. This would prevent the electrostatic field of the generator from spreading out of the cage. vt (ii) A high voltage generator is usually enclosed in such a cage which is earthen. (i) It is for this reason that it is safer to sit in a car or a bus during lightening rather than to stand under a tree or on the open ground. 40  Electrostatics: Part 1 . The electrostatic shielding can be achieved by protecting and enclosing the sensitive . di In g 17. When conductor is not earthed field of the charged body aP C due to electrostatic induction continues beyond AB. Sensitive instruments and appliances are affected seriously with strong external electrostatic fields.indd 40 3/31/2017 4:31:51 PM . If AB is earthed.2 Different Patterns of Electric Field Lines in a rn Le a ge ng Ce Module-Vol-II_01. Electrostatic shielding: Electrostatic shielding/screening is the phenomenon of protecting a certain region of space from external td electric field.L instruments inside a hollow conductor because inside hollow conductors. electric fields is zero. L vt aP Neutral point (N) is the location where the net electric field due to charges is zero. Le a ge ng Ce Module-Vol-II_01. td . It lies near the charge of smaller di magnitude. In g in a rn The figure shows the sketch of field lines for two point charges 2Q and –Q.indd 41 3/31/2017 4:31:51 PM . Electrostatics: Part 1  41 . (iv) The electric field lines should not form closed loops which they do as in Fig. Module-Vol-II_01. (d). the lines should be radially outward. i. Therefore. vt Neutral point lies near the position of smaller magnitude charge. the pattern must be symmetrical about the line joining the two charges. the lines are radial and spherically symmetric. No lines should pass through this point.indd 42 3/31/2017 4:31:51 PM . Number of lines: Twice as many lines leave +2Q as enter –Q.e.. (a).  Illustration 29    Which among the curves shown in figure cannot possibly represent electrostatic field lines? [NCERT] g in a rn Le a ge ng Solution Figures (a). they are continuous curves without breaks. Therefore. (b). there is an equivalent point below it. Why not?  [NCERT] di Solution The electric field due to a charge decreases continuously with distance and becomes zero only at infinity. Far field: Far from the system of charges. 42  Electrostatics: Part 1 .  Illustration 28    An electrostatic field line is a continuous curve. Ce (iii) The electric field lines should not intersect each other which they do as in Fig.L Near field: Very close to a charge. The pattern of field lines can be deduced by considering the following points: td Symmetry: For every point above the line joining the two charges. . Null point or Neutral point: There is one point at which E = 0. its field predominates. (ii) The electric field lines should start from positive charge and end at negative charge which is not so in Fig. the pattern should look like that of a single point charge of value (2Q –Q) = +Q. a field line cannot have sudden breaks. Since electric field lines In represent the electric field. That is. (d) and (e) cannot represent electric field lines due to the following reasons: (i) The electric field lines should be normal to the conductor which is not so in Fig. (e). (b). aP Note: Excess lines from 2Q charge will meet at infinite. Unit: (i) The SI unit of electric flux is Nm2 C–1 (gauss) or J m C–1. Electrostatics: Part 1  43 . (It can be positive. The electric flux of the field   over the area element is given by df E = E ◊ dS . while inward flux is to be negative di In g in 18. Let us select a small area element dS on this surface. (c) is correct as it represents electric field lines due to two equal positive charges. then find the flux through the square.2 m2 parallel to the Y–Z plane. a (ii) Electric flux is a scalar quantity.0 × 103 N/C.1  Physical Meaning The electric flux through a surface inside an electric field represents density of the electric lines of force crossing the surface in a rn direction normal the surface. Solution ge   Ê 3  4 ˆ N-m 2 f E = E ◊ S = Á E0 i + E0 j ˜ ◊ (0. It is a property of electric field.) Le  3  4   Illustration 30    The electric field in a region is given by E = E0 i + E0 j with E0 = 2.indd 43 3/31/2017 4:31:52 PM .2 iˆ ) = 240 Ë5 5 ¯ C a  Illustration 31    A point charge Q is placed at the corner of a square of side a.   aP The electric flux over the whole area is given by fE = E. ng Ce Solution The electric field due to Q at any point of the square will be along the plane of square and the electric field line are perpendicular to square.L Consider some surface in an electric field E. so f = 0. Find the flux of this 5 5 field through a rectangular surface of area 0. Only Fig. Module-Vol-II_01.dS = En dS ÚS ÚS   If the electric field is uniform over that area then fE = E ◊ S For a closed body outward flux is taken to be positive. negative or zero. It is along n̂ vt or dfE = EdS cos q  or  dfE = (E cos q) dS  or  dfE = En dS  where En is the component of electric field in the direction of dS.  Direction of dS is normal to the surface. td 18.  Electric Flux   . Æ Æ a f = E ◊ A = EA cos 60∞ = (3 ¥ 103 ) (10 -2 ) cos 60∞ = 15 Nm 2 /C Le  Illustration 34    What is the net flux of the uniform electric field of previous problem through a cube of side 20 cm oriented so that its faces are parallel to the coordinate plane? [NCERT] Solution  ge → For four faces (top. td  Illustration 32    Find out flux through the curved surface of the hemisphere of radius R if it is placed in uniform electric field E as shown in figure. E and A are perpendicular to each other and as such electric flux linked with each of these faces..e.EA Flux linked with face 2.EA + EA = 0 Module-Vol-II_01. Æ Æ f = E◊A = 0 a Flux linked with face 1.. . we can say that no line is crossing the square so flux = 0. front and back). so. i. i. Æ E = 3 ¥ 103 iˆ N/C or E = 3 ¥ 103 N/C In g A = 10 cm × 10 cm = 100 cm2 = 10–2 m2 Æ Æ in (a) f = E ◊ A = EA cos 0∞ EA = (3 ¥ 103 ) (10 -2 ) = 30 Nm 2 /C rn Æ  (b) Since angle between E and A is 60°. f = EpR2  Illustration 33    Consider a uniform electric field E = 3 ¥ 103 iˆ N/C. In other words.indd 44 3/31/2017 4:31:53 PM .e. bottom. di (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the YZ plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the X-axis? [NCERT] Solution Here.e. i. 44  Electrostatics: Part 1 .L vt Solution aP The electric lines which are passing through area pR2 are also the same which will pass through hemisphere.. Æ Æ Ce f2 = E ◊ A = EA cos 0∞ = EA Net flux through the cube. ng Æ Æ f1 = E ◊ A = EA cos 180∞ = . fc = 4f + f1 + f2 = 0 . in a uniform electric field E.Eab sin q = 0 ∑ Note that the contribution to the flux for a closed surface is positive for the surface where the field is directed out and negative for the surface where the field is directed into the surface.L vt aP Solution Note that flux through the faces ABF. In all the three cases. length l. Electrostatics: Part 1  45 .q ) = Eab sin q rn Flux through face ADEF: Magnitude of area vector of face ADEF = (sin q) b = ab sin q f E = E cos 180∞ (ab sin q ) = . If the field is uniform. Module-Vol-II_01. f total = (f E ) ABF + (f E )CDE + (f E ) BCEF + (f E ) ABCD + (f E ) ADEF ge = 0 + 0 + 0 + Eab sin q . area vector of CDE points in the negative z-direction and area vector of BCEF points in the negative y-direction. Compute the electric flux if Ce the axis of the cylinder is parallel to the field direction.  Illustration 36    Consider a cylindrical surface of radius R. the ng number of lines that enter the closed surface equals the number of lines that come out.  Illustration 35    Find the flux of the electric field through each of the five surfaces of the inclined plane as shown in figure. therefore total flux is algebraic sum of flux through each surface. Area vector of face ABF points in the positive z-direction. di In g          Flux through face ABCD: in Magnitude of area vector of face ABCD = ab Æ Æ f E = E ◊ A = E ( ab) cos (90∞ . (f E ) ABF = 0. td What is the total flux through the entire closed surface? . (f E )CDE = 0 and (f E ) BCEF = 0 Le (f E ) ABCD = + Eab sin q .Eab sin q Flux is a scalar quantity.indd 45 3/31/2017 4:31:54 PM . field E is normal to area vector. ∑ The flux of a constant vector through any closed surface is zero. (f E ) ADEF = . a ∑ The net flux for this closed surface can also be seen to be zero from examination of the field lines. CDF and BCEF is zero.Eab sin q a Here. A surface on which Gauss’s law is applied.   (f E ) right end = Ú right end E ◊d A = Ú right end Ú E dA cos 0∞ = + E  right end dA = Ep R 2 Finally at every point on the curved surface the area vectors are perpendicular to the direction of the electric field. a ge ng Ce Module-Vol-II_01. Gauss’ law provides a different g way to express the relationship between electric charge and electric field. While completely equivalent to Coulomb’s law. This law is useful in calculating field caused by charge distributions that have various symmetry properties. Hence the surface integral consists of the sum of the three terms: .Ep R 2 Note that E is constant over the entire plane surface of left end.L         fE =  Ú E ◊d A = Ú left end E ◊d A + Ú right end E ◊d A + Ú curved E ◊d A All the area elements on the left end and electric field E are at an angle of 180°   vt Ú (f E )left end =  left end E ◊d A = Ú left end E dA cos 180∞ = .E  Ú left end dA = . we take it out from the integral.   q Ú E ◊ ds = ein0 Mathematically. Gauss’s law can be written as f =  rn We once again emphasise that the electric field appearing in the Gauss’s law is the resultant electric field due to all the charges present inside as well as outside the given closed surface. there need not be any material object at the position of the surface. the charges qin appearing a in the law is only the charges contained within the closed surface. Caution: Remember that the closed surface in Gauss’s law is imaginary. (f E )curved =  Úcurved surface E ◊d A =  Ú curved surface E dA (cos 90∞) = 0 Total flux = (f E ) right end + (f E )left end + (f E )curved surface = ( + Ep R 2 ) + ( . right and left plane faces and curved portion of its surface. in Then. Solution td We can divide the entire curved surface into three parts. is sometimes called the Gaussian surface. all the area elements on the right end are parallel to electric field. E = 0.indd 46 3/31/2017 4:31:55 PM . therefore. Gauss’s Law In Gauss’s law is an alternative to Coulomb’s law. We often refer to a closed surface used in Gauss’ law as a Gaussian surface. On the other hand. 46  Electrostatics: Part 1 .   di Thus.Ep R 2 ) + 0 = 0 19. aP Similarly. Gauss’s law states that the total electric flux through a closed surface is proportional to the total electric charge enclosed with in the surface. Le The contribution of the charges outside the closed surface in producing the flux is zero. Consider the two Gaussian surfaces A1 and A2 as shown in figure. For surface A2. whereas Qnet is the charge enclosed by the Gaussian surface. we can use Gauss’ law to find the magnitude of E .  Problem Solving Strategy 20.L vt aP Note that the electric field in the expression Ú E. dA is the resultant field on the Gaussian surface. di Charge Q lies at the centre of the Gaussian surface A1. Electrostatics: Part 1  47 . g in a rn Le 20. In either case. then that point must lie on your Gaussian surface.1  Identification of the Relevant Concepts Gauss’s law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly ge Æ over a plane. charge Q is outside A2. so that the net flux through e0 In A2 is zero. If you are trying to find the field at a particular point. Ce Charge distribution Gaussian surface Electric field Point charge Spherical Radial Spherical charge distribution Spherical Radial Line of charge Cylindrical Radial Planer charge Plane parallel to charge distribution Normal to surface Module-Vol-II_01. For surface A1 the net flux Q through A1 is . begin your analysis by asking the question.indd 47 3/31/2017 4:31:55 PM . In these situations we determine the direction of E from the symmetry of the charge distribution. We often call it a Gaussian surface. Note that the field lines that enter the Gaussian surface (net flux in) also leave it (net flux out). we can use Gauss’s law to determine the details of the charge distribution. “What is a the symmetry?” ng The problem is solved by using the following steps: (a) Select the surface that you will use with Gauss’s law. Alternatively. if we are given the field. If we are given Æ the charge distribution. td . 07 ¥ 10 C = 0. (a) If the radius of the Gaussian surface were doubled. it may be in empty space. could you conclude that there were no charges inside the aP box? Why or why not? [NCERT] Solution 3 2 Here: fc = 8. ng 20.2  Field of a Charged Conducting of Sphere We place positive charge q on a solid conducting sphere with radius R (as shown in figure). where q is the net charge inside the box.  Illustration 38    A point charge causes an electric flux of –1. embedded in a solid body.85 ¥ 10 -12 C2 /Nm 2 ) ( -1. q remains Le Œ0 the same and so is fc. If the charge distribution has cylindrical or spherical symmetry. we take as our Gaussian surface an imaginary sphere of radius r centred on the conductor. Ce Module-Vol-II_01.L surface to be a coaxial cylinder or a concentric sphere. choose the Gaussian . Hence.0 ¥ 103 Nm 2 /C r = 10.8.0 cm radius centred on the charge. how much flux would pass through the surface? in (b) What is the value of the point charge? [NCERT] Solution rn Here. fc = –1. there may be charges inside the box whose algebraic sum is zero. (b) The Gaussian surface does not have to be a real physical surface.0 cm = 10 × 10–2 m = 10–1 m a q (a) As fc = .85 ¥ 10 -12 ˜ Á 8. q = 0.0 ¥ 10 Nm /c di q Ê C2 ˆ Ê 3 Nm 2ˆ (a) As fc = . (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero.  Illustration 37    Careful measurement of the electric field at the surface of a black box indicates that the net outward flux vt through the surface of the box is 8.85 ¥ 10 -9 C = .e.0 × 103 Nm2/C. such as a surface of a solid body.0 ¥ 103 Nm 2 /C) = .0 × 103 Nm2 / C (b) q = Œ0 fc = (8. Often the appropriate td surface is an imaginary geometric surface.8. q = Œ0 fc = Á 8. or both.0 × 103 Nm2/C to pass through a spherical Gaussian surface of g 10. (c) Usually you can evaluate the integral in Gauss’ law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. it is clear that fc depends on q only.1.. i. 48  Electrostatics: Part 1 .85 nC ge 20. to  calculate field inside. In either case. we take r to be greater than the conductor’s radius R. All the charges must be on the surface a of the sphere.07 mC Œ0 Ë Nm 2 ¯ Ë C ¯ In (b) When fc = 0. fc = .0 ¥ 10 -6 ˜ = 0.indd 48 3/31/2017 4:31:56 PM . To take advantage of the symmetry. the point where we want to calculate E lies on Gaussian surface. we take r to be less than R. even when the radius of the Gaussian surface is doubled. To calculate the field outside the conductor. we can only conclude that the net charge inside the box is zero. Thus. respectively. Thus.3  Selection of Gaussian Surface The system has spherical symmetry. height l. The area of the Gaussian surface is 4pr2. This property suggests that we use as a ge Gaussian surface a cylinder with arbitrary radius r and arbitrary length l. Module-Vol-II_01. Ú E^ dA in Gauss’s law is therefore just E(4pr ) gives 2 .  a There is no flux through the ends because E lies in the plane of the surface. From Gauss’s law. 20.4. (To make a paper cylinder with radius r and height l.4  Electric Field Outside the Sphere td We first consider the field outside the conductor.L The flux integral vt aP di In g in q 1 q E (4p r 2 ) = and E = (outside a charged conducting sphere) e0 4pe 0 r 2 rn This expression for the field at any point outside the sphere (r > R) is the same as for a point charge. so we choose r > R. which is the charge per unit length multiplied by the length of wire inside the Gaussian surface. E is uniform over the surface and perpendicular to it at each point. note that E is perpendicular to the surface at each point. which is (E) (2prl) and the zero flux through the two ends. F E = ( E )(2p rl ) = l l and E = 1 l e0 2pe 0 r (field of an infinite line of charge). E has the same value everywhere on the walls. by symmetry. We break the surface integral for the flux FE into an integral over each flat end and one over the curved side walls. Electrostatics: Part 1  49 . Ce Finally. The charge per unit length is l (assumed positive). we need the total enclosed charge. where r = R. thin wire. so the  enclosed charge is q. The entire conductor is within the Gaussian surface. a 1 q E= (at the surface of a charged conducting sphere) 4pe 0 R 2 Le 20. Hence the total flux FE through the entire cylinder is the sum of the flux through the side walls.1 Field of a Line Charge Electric charge is distributed uniformly along an infinitely long. or Qenel = ll.indd 49 3/31/2017 4:31:57 PM . the field due to the charged sphere is the same as though the entire charge were concentrated at its centre. you need a paper rectangle with width 2pr.  ng To find the flux through the side walls. The area of the side walls is 2prl. with its ends perpendicular to the wire. Just outside the surface of the sphere. The system has cylindrical symmetry. and area 2prl). a One end face. if s is negative. vt The charged sheet passes through the middle of the cylinder’s length.) Hence the total flux through ge the surface is E^A. we use as our Gaussian surface a cylinder with its axis perpendicular to the sheet of charge. that is. Hence. E is directed radially inward toward the line of charge. 20. the flux through the Gaussian surface in figure is negative. lies within the conductor and the other lies just outside. we construct a Gaussian surface in the form of a small cylinder (as shown in figure). and the direction of  E is radial at every point on the surface. field is very nearly uniform and perpendicular to plane. 20. Because E is perpendicular aP to the charged sheet. the total electric flux through the Gaussian surface is the product of E and the total area of the surface A = 4pr2. The electric field is zero at all points within the conductor. Module-Vol-II_01. or Qenel = sA. The assumption that the sheet is infinitely large is an idealization. with area A. At such points.5. (If s is positive. it is parallel to the curved side walls of the cylinder. FE = 4pr2E. cylindrical and plane surfaces.6  Electric Field Inside the Sphere Ce From symmetry the magnitude E of the electric field has the same value at every point on the Gaussian surface. infinite sheet on which there is a uniform positive charge per unit area s. E is directed toward the sheet. If it is negative. The charge enclosed within the Gaussian surface is sA.indd 50 3/31/2017 4:31:57 PM . 50  Electrostatics: Part 1 . and in the above td expression for the field magnitude E we must interpret l as the magnitude (absolute value) of the charge per unit length. 20. The total flux integral in Gauss’s law is then 2EA (EA from each end and zero from the side walls). so from Gauss’s law. with ends of area A.  We have assumed that l is positive. The net charge within the Gaussian surface is the charge per unit area multiplied by the sheet area enclosed by the surface.5 Field at the Surface of a Conductor rn To find a relation between s at any point on the surface and the perpendicular component of the electric field at that point. Outside the conductor the Le  component of E perpendicular to the side walls of the cylinder is zero.L Let us consider a thin. Gauss’s law gives di sA s 2 EA = and E = (field of an infinite sheet of charge) e0 2e 0  If the charge density is negative. is  perpendicular to the surface. nothing in nature is really infinitely large. the flux through each end. and there is no flux through these walls.4. To take advantage of these symmetry properties. so the cylinder. the field points inward and E^ is negative. But the result g s E= is a good approximation for points that are close to the sheet (compared to the sheet’s dimensions) and not too near its 2e 0 in edges. flat. sA s E^ A = and E^ = (field at the surface of a conductor) e0 e0 a We can check this with the results we have obtained for spherical. the electric field points out of the conductor and E^ is positive. ng 20. and 2e 0 In in the expression E = s denotes the magnitude (absolute value) of the charge density.1 Field of a Uniformly Charged Sphere Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R. and over the end face the perpendicular component is equal to E^. Hence. hence.2 Field of an Infinite Plane Sheet of Charge . f = Ú E ◊ ds = e0 aP Q r3 1 Qr E Ú ds = E ◊ 4p r 2 = or E = (field inside a uniformly charged sphere) e0 R 3 4pe 0 R3 The field magnitude is proportional to the distance r of the field point from the centre of the sphere. Qencl = rVencl = Á ˜ Á pr ˜ = Q 3 3 4p R 3 Ë 4p R / 3 ¯ Ë 3 3 ¯ R vt Qenc.L sphere of radius R: Q Ê Q ˆ Ê 4 3ˆ r3 r= . The volume charge density r is the charge Q divided by volume of the entire charged . td Let’s first find the field magnitude inside the charged sphere of radius R. This surface encloses in the entire charged sphere.indd 51 3/31/2017 4:31:58 PM . Electric field in terms of charge density E= 4pe 0 Ê 4 1 ÁË 3 R 3 ˆ r ◊ p R3 ˜ r ¯ = rr 3e 0 fi E= rr 3e 0 In (field inside a uniformaly charged sphere) g To find the field magnitude outside the charged sphere. so we choose r < R. we use a spherical Gaussian surface of radius r > R. The amount of charge enclosed within the Gaussian surface depends on the radius r. di At the centre (r = 0). a ge ng Ce Module-Vol-II_01. E = 0. Then using Gauss’s law. so Qenel = Q and Gauss’s law gives Q 1 Q E ◊ 4p r 2 = or E = (field outside a uniformly charged sphere) e0 4pe 0 r 2 rn For any spherically symmetric charged body the electric field outside the body is the same as though the entire charges were concentrated at the centre. Electrostatics: Part 1  51 .7  Electric Field Due to a Long Uniformally Charged Cylinder Le Consider a long uniformally charge cylinder of volumetric charge density r and radius R. a 20. the magnitude E is evaluated at the radius of the Gaussian surface. a The electric field at the surface of the inside sphere goes to zero after connection. For any point r < R or r > R the Gaussian surface will be cylindrical as shown in figure. how much total charge is on the outer sphere? How much charge is on the outer surface and inner surface of outer sphere? Does the electric field at the rn surface of the inside sphere change when the wire is connected? We return to original condition in (a). Total charge on inner surface is 0. in If a wire is connected between the inner and outer spheres. The final charge distribution is shown in figure. In Ë r¯  Illustration 39    Consider two concentric conducting spheres. Total charge on outer surface of outer sphere is –5Q. td For any point inside the cylinder (r < R). an equal amount of charge must come on the outer surface. Qenclosed f E = E (4p r 2 ) = = 0. on Le the inner surface of outer sphere. (c) When the outer sphere is grounded the charge on this surface is transferred to ground. we have E = 0. Sum of all the induced charged is always zero. qin ( r ◊ p r 2 l ) E ◊ (2p r l ) = = . ng Consider a Gaussian surface just on the surface of inner sphere. In electrostatic equilibrium charge does not reside inside a conductor. We now connect the outer sphere to ground will be on the outer sphere? How much charge will be on the inner surface and outer surface of the outer sphere? a Solution (a) The charge on the inner sphere induce equal magnitude of charge. as Qenclosed = 0 e0 Ce Thus. the entire charge is transferred to ge the outer sphere from inner sphere. thus charge is reduced to zero. Module-Vol-II_01. How much charge is on the outer surface and inner surface of the outer sphere. (b) When outer and inner spheres are connected by a wire. Therefore.indd 52 3/31/2017 4:31:59 PM .L e0 e0 rr E= fi E µr 2e 0 vt For any point outside the cylinder (r > R) qin ( r ◊ p R 2 l ) E ◊ (2p r l ) = = aP e0 e0 r R2 1 E= fi Eµ 2e 0 r r di Electric field inside the long uniformally charged cylinder varies linearly E µ r and outside the cylinder the electric field varies inversely proportional to the distance Ê 1ˆ from the axis Á E µ ˜ . g The inner sphere is solid and has a charge +2Q on it. but opposite in sign. 52  Electrostatics: Part 1 . after electrostatic equilibrium is established. The outer sphere is hollow and initially has a charge –7Q on it. Thus outer and inner surface of outer sphere have charges –5Q and –2Q respectively. e0 rn • All the six surfaces are symmetrical with respect to charge. Hence.indd 53 3/31/2017 4:31:59 PM .L The faces adfe.al 2 As the field at the face efgh (that lies in the yz plane. y = 0) is E = ajˆ and area vector is aP l 2 ( . 6 6e 0 a Le  Illustration 42    In figure shown a charge q is placed at a distance d Æ 0 near one of the edges of a cube of edge l on a line of symmetry along diagonal.   Illustration 40    A cube of side l has one corner at the origin of coordinates and extends along the positive x. y and z-axes. Solution g • This problem can solved by symmetry consideration and Gauss law. vt Æ Æ f2 = Ú E ◊ dA = a( ˆj ) ◊ l 2 ( . q • The flux enclosed with the cube f = . • We can enclose the charged particle by a cube of side ‘a’ and keeping the particle at the centre of the in cube.ˆj ) (direction outward normal) Flux through face abcd f = (a + bl ) ˆj ◊ l 2 ˆj = (al 2 + bl 3 ) Net flux through the cube = f1 + f2 = bl3 Qenclosed di From Gauss’s law f = . adhe will contribute zero flux because the area vector is normal to electric field. What is the flux through the other three faces? a ge ng Solution • Use of symmetry consideration may be useful in problems of flux calculation. bcgf. Flux through face efgh. cdgh. Solution . td Suppose the electric field in this region is given by E = (a + by) ˆj . q • The total flux passing through the close cube f = . e0 Module-Vol-II_01. What is flux through each of the sides containing the point a. Qenclosed = e 0f E = e 0 bl 3 e0 a In  Illustration 41    A point charge q is placed at a distance from the centre of a square of side ‘a’ as shown in the diagram. 2 calculate the electric flux passing through the square. Determine the charge inside the cube. hence they will have equal contribution f q of the flux f ¢ = = . Ce • We can imagine a charged particle is placed at the centre of a cube of side 2l. Electrostatics: Part 1  53 .ˆj ) = . 3 24e 0  Illustration 43    Two identical metal plates each having surface area ‘A’.L vt aP f q • The flux passing through one of the face of the cube. q a • Hence same amount of the flux will pass through the sides contain the point a . 24e 0 8e 0 • As d Æ 0 we can say the faces (abcd). 54  Electrostatics: Part 1 . (2). 8e 0 q Le • The electric field lines are towards the faces containing the point a hence the flux will be negative i. i. rn • The number of electric field lines which are passing through the sides which do not contain the point a are same as the number of the electric field lines passing through the sides containing the point a. (bcfg) and (dchg) are away from the faces hence the flux associated with each of the faces will be positive q ˆ ÁË i.e. 4 24e 0 di • Each of the face (efgh). 8e 0 f ¢¢ q • Hence the flux through each of the faces containing the point ‘a’ will be =. Find the charge appearing on surface (1). f ¢¢ = . Assume the size of the plate is much longer than the separation between the plates.e. . a ng Ce Module-Vol-II_01. g 0 3¥q in q • Hence total flux through these side = + . having charge ‘q1’ and ‘q2’ are placed facing each ge other at a separation ‘d’. . .indd 54 3/31/2017 4:32:00 PM . td . + 24e ˜¯ . (3) and (4). 24e 0 Ê In • The electric field lines for the faces (efgh). 6 6e 0 f¢ q • Hence the flux passing through the face bcfg = = . (bcfg) and (dchg) are symmetrical with respect to q change hence the flux passing through each of the face is . (abef) and (adeh) are also symmetrical about charge. f ¢ = = .e. indd 55 3/31/2017 4:32:01 PM . Let us consider a point ‘P’ inside the left plate. Let the facing surfaces have the charge x and –x [surfaces (2) and (3) respectively].q 2 ( q1 . vt aP di Æ (q1 .2 2e 0 2e 0 2e 0 2e 0 2 Ae 0 2 Ae 0 2 Ae 0 2 Ae 0 rn q1 . Then the charge on the surfaces (1) and (2) should be (q1 – x) and x respectively (by conservation of charge). Solution td Facing surfaces have equal and opposite charge (By Gauss theorem).( q 2 + x ) = 0 fi = 2 a Hence charge appearing on different surfaces are as follows: Le 1 2 3 4 (q1 + q2) (q1 – q2) (q1 + q2) 2 2 2 –(q1 – q2) ge (q1 + q2) Face (1) 2 2 Face (2) (q1 – q2) 2 a (q1 – q2) ng Face (3) – 2 Face (4) (q1 + q2) 2 Ce Module-Vol-II_01.L have equal and opposite charge hence charge appearing on surfaces (3) and (4) will be –x and (q2 + x) respectively. + .x ) .x ) Æ Æ Æ | E | = | E1 + E2 + E3 + E4 | = 0  In Net electric field at ‘P’ will be due to the resultant of electric field due to charge appearing on all four surfaces. Electrostatics: Part 1  55 . Facing metallic surfaces always .4 = 1 . | s 2 | = | s 3| = and s 4 = 2 A A A in Then from (i) Æ s1 s s s (q . Net electric field at ‘P’ should be zero. Æ (q + x ) …(i) g x s1 = .x) x x (q + x) | E |= 0 = .2 + 3 .x + x . 56  Electrostatics: Part 1 . g 21. total solid angle around a point in space is the solid angle subtended by entire spherical surface on its centre. If the conducting body is connected to earth have any charged object near to it the body will not have zero charge.. Note: td • Facing surfaces have the equal and opposite nature charge with magnitude ‘half the difference of the charge on different Êq -q ˆ (q .e.e. The magnitude of solid angle subtended by an area S at a point is define as rn cos q W=Ú ds S r2 a at its centre. Ë 2 ¯ 2 Ë 2 ˜¯ . If conducting body is isolated and connected to earth it will have no charge.L (q + q2 ) • Outer surfaces always have equal charges of magnitude ‘half the summation of charges. ˜ fi W = 2p Á1 ..Á 1 in surface (3). • The charge appearing on the surfaces (2) and (3) is called bounded charge and the charge appearing on the surfaces (1) and vt (4) is called free charge. 2 • If we have this type of charge distribution then the electric field inside any metal plate will be zero. 1 in each surfaces (1) and (4). i. Le 4p R 2 W0 = = 4p steradian R2 Solid angle subtended by a disk at a point on its axis: Consider a coaxial area element of radius x and thickness dx.q2 ˆ plates’. where a is the semi vertical angle of the cone subtended by the disk at P. • If we join second plate (right plate) with ground the charge appearing on the surface (4) will go to the earth and charge distribution will be aP di In • Any metal plate or object connected to the earth need not to have zero charge. Appendix in Solid angle: It is the cone subtended by an area at the point of interest. dS = 2p xdx ge Solid angle subtended by this element at point P is dS cos q 2p xdxa dW = 2 2 fi dW = (x + a ) ( x + a2 ) a2 + x 2 2 a Hence total solid angle subtended by the disk is ng R R 2 xdx Ê 1 ˆ Ê a ˆ W = p aÚ fi W = 2p a Á . ˜ 0 ( x 2 + a 2 )3/2 Ë a2 + x 2 ¯ 0 Ë a2 + R2 ¯ Ce W = 2p (1– cos a). i.indd 56 3/31/2017 4:32:02 PM . Module-Vol-II_01.q ) Ê q . Therefore. Á 1 2 ˜ in surface (2) and 2 1 or . A charge q is placed at the centre of a cube. Electrostatics: Part 1  57 .cos q ) q(1 .L Solution Method 1: For point charge Gaussian surface should be spherical. 2e 0 . Solid angle subtended by the base of the cone at the apex of cone is p = 2p (1 – cos q).cos q ) q The desired flux is f E = Á ˜ ◊ = ◊ = Ë A0 ¯ e 0 (4p R 2 ) e0 2e 0 Method 2: Using the concept of solid angle. As the in q flux associated with solid angle 4p is . The inward and outward electric flux for a closed surface in units of N-m2/C are respectively 8 × 103 and 4 × 103. dA = (2pr)Rda = (2pR sin a) R sin a) Rda as r = R sin a = (2pR2) sin a da q di A = Ú (2p R 2 )sin a da . Consider a differential ring of radius r and thickness dr. Total slid angle around a point in space is 4l steradian. Then the flux passing through one face of cube will be q q (a) (b) a e0 2e 0 ng q q (c) (d) 4e 0 6e 0 3. g Solid angle is the cone subtended by an arc at the point of intersect.indd 57 3/31/2017 4:32:03 PM .cos q ) the base of the cone is . The flux through the whole sphere is q/e0. A = 2p R 2 (1 . If a spherical conductor comes out from the closed surface of the sphere then total flux emitted from the surface will be Ce 1 (a) ¥ (the charge enclosed by surface) (b)  e0 × (charge enclosed by surface) e0 1 (c) ¥ (charge enclosed by surface) (d)  0 4pe 0 Module-Vol-II_01.  Illustration 44    A point charge q is placed on the apex of a cone of semi-vertex angle q.cos q ) is f = = e0 4p 2e o a Concept Application Exercise 4 Le 1. A0 = area of whole sphere = 4pR2 aP and A = area of sphere below the base of the cone. Therefore. Here. e0 rn q 2p (1 .cos q ) 0 In Ê A ˆ q (2p R 2 ) (1 . Then the total charge inside the surface is [where e0 = permittivity constant] (a) 4 × 103 C (b) –4 × 103 C 3 (c) ( -4 ¥ 10 ) C (d) –4 × 103 e0 C ge e 2. the flux through the base of the Ê Aˆ q fE = Á ˜ ◊ vt Ë A0 ¯ e 0 cone. Show that the electric flux through td q (1 .cos q ) q (1 .cos q ) Hence the flux associated with solid angle 2p (1 . Consider a Gaussian sphere with its centre at the apex and radius the slant length of the cone. When calculating the flux of the di electric field over the spherical surface the electric field will be due to In g (a) q2 (b) Only the positive charges in (c) All the charges (d) +q1 and –q2 7. Shown below is a distribution of charges. The flux of electric field due to these charges through the surface S is vt aP (a) 3q/e0 (b) 2q/e0 (c) q/e0 (d) Zero 6. A charge q is located at the centre of a cube. the electric field is directly proportional to Module-Vol-II_01.L q 2p q (c) (d) 6(4pe 0 ) 6(4pe 0 ) 5. 4.32 × 103 Nm2 C–1 (d) 6. The electric intensity due to an infinite cylinder of radius R and having charge q per unit length at a distance r(r > R) from its axis is (a) Directly proportional to r2 (b) Directly proportional to r2 (c) Inversely proportional to r (d) Inversely proportional to r2 Ce 10. Consider the charge configuration and spherical Gaussian surface as shown in the figure. for x < R.32 × 103 CN–1 m–2 ng 9. q2 = 78. Which statement is correct? (a) Electric flux is coming towards sphere rn (b) Electric flux is coming out of sphere (c) Electric flux entering into sphere and leaving the sphere are same a (d) Water does not permit electric flux to enter into sphere 8.85 nC. Le q3 = – 56 nC) ge (a) 103 Nm2 C–1 (b) 103 CN–1 m–2 a (c) 6. The electric flux for Gaussian surface A that enclose the charged particles in free space is (given q1 = –14 nC.indd 58 3/31/2017 4:32:03 PM . An electric dipole is put in north-south direction in a sphere filled with water. At a distance x from its centre. A sphere of radius R has a uniform distribution of electric charge in its volume. 58  Electrostatics: Part 1 . The electric flux through any face is td 4p q pq (a) (b) 6(4pe 0 ) 6(4pe 0 ) . There exists an electric field given  by E = ÈÎ(2. where x is in metres.0 m. The net charge in coulombs enclosed by the cube is equal to vt aP (a) –54 Œ0 (b) 6 Œ0 (c) –6 Œ0 (d) 54 Œ0 di 12. td 1 1 (a) (b) 2 x x (c) x (d) x2 . The variation of electric field as a function at x (for x = 0 to x = 3d) will be best represented by In g in rn (a) (b) a Le ge (c) (d) 13.0 k ˘˚ N/C. Electrostatics: Part 1  59 . in the region in which it lies. A cylindrical portion of radius r is removed from a solid sphere of radius R and uniform volume charge density in such a way that the axis of the hollow cylinder coincides with one of the diameters of the sphere (r is negligible compared to R).L 11. They are carrying charges ‘Q’ and ‘3Q’ respectively. Two very large thin conducting plates having same cross-sectional area are placed as shown in figure.0) i + 8.0 j + 3. Figure above shows a closed Gaussian surface in the shape of a cube of edge length 3.0 x + 4.indd 59 3/31/2017 4:32:04 PM . Then the electric field intensity at point A is a ng Ce Module-Vol-II_01. i 3e 0 3e 0 rr ˆ rr ˆ . td rr ˆ rr ˆ (a) i (b) . Three uncharged conducing large plates are placed parallel to each other in a uniform electric field. x = +a. Find the induced charge density on each surface of each plate. y = +a. Find the electric flux crossing out of the cube in the unit of 10–4 Nm2 /C. A system consists of a ball of radius R carrying a spherically symmetric charge and the surrounding space filled with a charge of volume density ρ = α/r. where α is a constant. z = 0  and z = +a. y = 0. Find the ball’s charge for which the magnitude of the electric field strength vector is independent of r outside the ball.1 m placed such that its six faces are given by equations x = 0. How large is this strength? The primitivities of the ball and the surrounding space are assumed to be equal to unity.indd 60 3/31/2017 4:32:04 PM . 60  Electrostatics: Part 1 .L (c) i (d) . vt aP di 15. placed in electric field given by E = x 2iˆ + y ˆj N/C. Consider a cube of side a = 0. i 6e 0 6e 0 14. ‘r’ is the distance from the centre of the ball. g in a rn Le a ge ng Ce Module-Vol-II_01. In 16. L More than One Concepts qC = 2 mC and qD = –5 mC are located at the corners of a square ABCD of side 10 cm. SOLVED EXAMPLES td Miscellaneous Problems Based on One or   Example 3    Four point charges qA = 2 mC. What does the ratio signify? [NCERT] r2 a Solution ke = [2rˆOA . If n is the number of electrons transferred.6 ¥ 10 -19 )2 of each string is l and the angle is q. Fe. The sphere is in equilibrium.rˆOA and rˆOD = .rˆOB = 0. Æ È1 ¥ 2 1( -5) 1¥ 2 1( -5) ˘ ke 2 Thus. repulsive electric force. C to O and D to O. F = ke Í 2 rˆOA + 2 rˆOB + 2 rˆOC + 2 rˆOD ˙ rn   Example 2    Check that the ratio is dimensionless. . i.77 ¥ 10 -11 )(9. each having ge a mass m. qB = –5 mC. force ng force (Fg) between an electron and a proton at the same distance of gravity. = e 1 2 2 = e     (as q1 = q2 = e)  Fx = T sin q .indd 61 3/31/2017 4:32:05 PM .2 rˆOA + 5 rˆOB ] Le È ke e ˘ 2 2 2 2 r2 [Nm /C ][C ] Í ˙= 2 2 = no units  ÍÎ Gme m p ˙˚ [Nm /kg ][kg][kg] (as rˆOC = . respectively.Fe = 0  Ce Fg Gme m p / r Gme m p …(i) This large ratio suggests that Fe >> Fg. The forces in the horizontal and vertical directions must separately Fe k q q / r2 k e2 add up to zero.  Fy = T cos q . = Gme m p (6. = (1. F = 0) As the ratio has no units.   Example 4    Two identical small charged spheres.1 ¥ 10 -31 )(1. Solution then Applying the principle of super-position. in Let OA = OB = OC = OD = r (say) Of course.6 ¥ 10 -27 ) charge on each sphere. since body diagram of the sphere. What is the force on a charge of 1 mC   Example 1    A polythene piece rubbed with wool is found to placed at the centre of the square? [NCERT] vt have negative charge of 3 × 10–7 C. Find the magnitude of the Further. it is dimensionless.875 ¥ 1012 Æ qq qq qq e 1. this mass relativity small.4 ¥ 1039 Solution a This is the ratio of the electric field (Fe) and the gravitational The forces acting on the sphere are tension in the string T. electrons are transferred from wool to polythene.5 rˆOD ˘˚ of this ratio.mg = 0  …(ii) Module-Vol-II_01. as shown in the free (r) apart. rˆOB . (a) Estimate the number of electrons transferred (from which to which)? aP (b) Is there a transfer of mass from wool to polythene?  [NCERT] Solution (a) Since the polythene piece acquires a negative charge (q) di on rubbing with wool.1 ¥ 10 -31 kg) = 1..875 ¥ 1012 ) (9.e. Electrostatics: Part 1  61 . rˆOC and rˆOD represent unit vectors from A to O. electrostatic force on In q0 (1 mC) placed at O (centre of the square) is given by q 3 ¥ 10 -7 C Èq q ˘ q = ne or n = = = 1. hang in equilibrium as shown in figure. The length ke e 2 (9 ¥ 109 ) (1. Î r r r r ˚ Gme m p Look up a Table of Physical Constants and determine the value ke = ÈÎ2rˆOA .7 ¥ 10 -18 kg B to O. ª 2. mg.5 rˆOB .6 ¥ 10 -19 C F = ke Í 02 A rˆOA + 02 B rˆOB + 02 C rˆOC + 02 D rˆOD ˙ ÍÎ rOA rOB rOC rOD ˙˚ g (b) Mass transferred from wool to polythene where rˆOA .5 rˆOB + 2 rˆOC . 5 × 103 N/C and point radially inward. E = 1. Find the dielectric r 2 ke in constant of the liquid 1Ê 1 ˆ or l= (2 ¥ 10 -2 m) (9 ¥ 10 4 N/C) 2 Ë 9 ¥ 10 Nm /C ˜¯ Á 9 2 2 Solution rn When the balls are in air. The density of liquid is s. q = – 6. l= does not change. the Coulombic (a) E in region I is zero. from which we obtain Fe = mg tan q Solution .s vg r . the charge on the sphere 4pe 0 (2l sin q ) di is negative. 1 |q| Er 2 4pe 0 As E= . Calculate the linear charge density.1 µC/m F = ( rvg) tan q  …(i)   Example 8    Two large. R = 10 cm = 10–1 m 2 = mg tan q  …(iii) r r = 20 cm = 2 × 10–1 m 1 where k= and r = 2 L sin q . what is td From equation (ii). angle q (b) E in region II is zero.s Ce   Example 6    A uniformly charged sphere of radius 10 cm has an unknown charge. Thus we can eliminate T from cos q the net charge on the sphere? [NCERT] equation (i). mg of the sphere is 1. On their inner faces. are suspended from a common point by two insulating strings Solution of equal length. E = 9 × 104 N/C. i.5 ¥ 103 N/C)(2 ¥ 10 -1 m)2 = aP 9 ¥ 109 Nm 2 /C2 -9   = 6.67 mC. What is E: (a) in the outer region of the first plate.L kq 2 Here. or q = 16pe 0 l 2 mg tan q sin 2 q   Example 7    An infinite line charge produces a field of 9 × 104 In   Example 5    Two identical balls each having a density r N/C at a distance of 2 cm. Both the balls have equal mass and charge. s = 17. 62  Electrostatics: Part 1 . If the electric field 20 cm from the centre Module-Vol-II_01. In Here. This is due to the reason that E exists only in the region a F ¢ = ( rvg .. thin metal plates are parallel and close to each other. is unchanged. W¢ = (rvg – svg).e. Now 2 ke l rE g both the balls are immersed in a liquid. According to the problem. T = .0 × 10–22 C/m2 ge When the balls are suspended in a liquid.67 ¥ 10 C = 6.s vg) tan q  …(ii) between the plates. | q | = vt 4pe 0 r 2 (9 ¥ 109 Nm 2 /C2 ) (1. r = 2 cm = 2 × 10–2 m equilibrium each string makes an angle q with the vertical.70 × 10–22 C/m2. (b) in Le the outer region of the second plate and (c) between the plates?  [NCERT] Solution Here. ng From equations (i) and (ii). As a result the angle q As E= . from previous problem.indd 62 3/31/2017 4:32:06 PM .67 nC 1 q2 Æ Equation (iii) now reduces to = mg tan q Since E points radially inwards. the plates have surface a charge densities of opposite signs and of magnitude 1.5 × 103 N/C. force is reduced to F¢ = F/K and apparent weight = weight – upthrust. we get (c) E in the region III is given by F rvg r =K= = F' rvg .   = 10–7 C/m = 0. t = ng vx Solution Vertical deflection of the particle at the far edge of the plate. i.. in Y-direction = 0] Module-Vol-II_01.e.e.85 ¥ 10 -12 C2 /Nm 2 .92 ¥ 10 -10 N/C e 0 8.e. Since inside the conductor.L   Example 9    A conductor A with a cavity as shown in figure (a) is given a charge Q. The length of plate is L and a uniform net charge Ú E◊ dS = fE =  =0 in electric field E is maintained between the plates. i. and s is the surface charge density near the hole. ( E 1 + E 2 ) + ( E 1 . Show that the e0 vertical deflecting of the particle at the far edge of the plate is This means that a charge (–q) must be present on the qEL2 . it should be enclosed in a metallic case where electric field is zero..   Example 10    A hollow charged conductor has a tiny hole ge cut into its surface.e. s Hence. where n̂ is the unit vector in the outward normal Fy qE 0 Acceleration experienced by the particle. 2 e0 Solution (a) Consider a Gaussian surface inside the conductor but close   Example 11    A particle of mass m and charge (–q) enters the g to the cavity.E 2 ) = di e0 (b) A sensitive instrument is to be shielded from the strong s electrostatic fields in its environment. Electrostatics: Part 1  63 . A just outside and B within the i. an equal and opposite charge (+q) must appear on the outer surface Solution of A. i. From equations (i) and (ii). Suggest a possible or E1 = 2 e0 [NCERT] In way. Fy = qE Le (b) For electrostatic shielding of the sensitive instrument. the electric field in the hole is . s 17. Let E1 be the electric field at A and B due to the entire 2 1 1 Ê qE ˆ Ê L ˆ shell except the material originally filling the cavity and E2 be y = ay t 2 = Á ˜ Á ˜ [as vy0. 2 2 Ë m ¯ Ë vx ¯ If there is no cavity. Show that the electric field in the hole is Ê s ˆ ÁË 2e ˜¯ nˆ . Show that the total s charge on the outside surface of A is Q + q. E = 0..0 ¥ 10 -22 C/m 2 td E= = = 1. [NCERT] rn inner surface of conductor A so that net charge inside the (2 mvx2 ) Gaussian surface is zero. This makes the total charge on the outer surface a Force acting on particle of charge –q in positive Y-direction. vt s E1 + E 2 =  (i) e0 Let us now consider the situation with cavity..e.. region between the two charged plates initially moving along Æ Æ X-axis with speed vx. (Q + q). E1 and E2 must cancel each other and as such. i. electric field at A. (a) Another conductor B with charge q is inserted into the E1 – E2 = 0 (ii) cavity keeping B insulated from A. m m a L  [NCERT] Time taken by the particle to cover the distance L. Since there aP is on field inside a conductor. at B. the initial velocity Ce its corresponding value for this material. To conserve charge. ay = = direction. Let A and B be two points.indd 63 3/31/2017 4:32:07 PM . cavity. Determine the ratio of the force Coefficient of friction between the particle B and the slope is Module-Vol-II_01. The disk slides down the dome without friction and placed on the slope at angular position b measured from the line breaks away from the surface of the dome at angular position of greatest slope passing through the position of the first particle.0 ¥ 10 ) Hence mgR ÊÁ 1 ˆ˜ + qER ÊÁ 3 ˆ˜ = mv When.2 m 4mg . a small   Example 15    A charge particle A is fixed at the base of a charged disk is gently released on the top of a fixed spherical uniform slope of inclination a.1 × 10–31 kg In eEL2 As q = e and m = me fi y = 2 me vx2 2 2 yme vx2 -2 2 (0.854 ¥ 10 -12   Example 14    In a horizontal uniform electric field. the path followed by the Ë 2 mvx ¯ charged particle is a parabola which is similar to the motion of a projectile in the gravitational field. If E between the plates separated by 0.1 × 10 –31 kg) [NCERT] 3 4 Given sin q = and cos q = 5 5 Solution y = 0. 64  Electrostatics: Part 1 .0 µC/m2.cos q ) + qER sin q = aP will the electron strike the upper plate? 2 (|e| = 1. R = = 1.1 ¥ 10 -31 6 2 )(2.qE sin q = rn 2. sin q = fi cos q = 5 5 Le Solution fi Ê 4ˆ Ê 3 ˆ mv 2 mg Á ˜ .6 × 10–2 m = 1.1 ¥ 10 2 ) g fi 5 mv 2  …(i)   = 2. conserving energy at A and B is an electron projected with velocity vx = 2.3qE a (b) Total electric flux leaving the surface of the sphere.1 × 102 N/C. q = sin–1(3/5) from the vertical.L As y µ L Á 2 being a constant ˜ .qE Á ˜ = Here.2)2 = 1. ( mg + 3qE ) 1 = (4mg . vx = 2.indd 64 3/31/2017 4:32:09 PM .4 mv 2  …(ii) Radius of the charged sphere.6 × 10–19 C.0 mC m -2 = 80 ¥ 10 -6 C m –2 Ë 5¯ Ë 5¯ R 2.1 × 102 N/C. me = 9. e = 1. R (a) Find the charge on the sphere. When the particle lose contact (b) What is the total electric flux leaving the surface of the 3 4 a sphere? [NCERT] N = 0.5 cm is 9.N . s = 80. where mv 2 mgR (1 . L2 = = Ë 5¯ Ë 5¯ 2 eE (1. Another charge particle B is Ce dome.3qE ) 2 = 80 ¥ 10 -6 ¥ 4p ¥ (1. E = 9.3qE = 5 2 R ge (a) Charge on the sphere.45 ¥ 10 -3 2 mg = 9qE fi = f= = = 1.5 cm = 0.0 × 106 m/s di Here. vt Solution   Example 12    Suppose that the particle in previous problem Motion from A to B.64 ¥ 108 N m 2 C-1 (qE ) 2 ng e 0 8.6 ¥ 10 -19 )(9.5 × 10–4 ( mg + 3qE ) = in 2 R or L = 1. of gravity acting on the disk to the force of its interaction with qEL2 td or y= the field.   Example 13    A uniformly charged conducting sphere of mv 2 mg cos q .6 cm Also equation of circular motion. 2 mvx2 2 Ê qE ˆ . me = 9. q = s × 4p R2 From (i) and (ii). ( mg) 9 q 1.0 × 106 m/s.5 × 10–2 m.5 ¥ 10 )(9.6 × 10 –19 C.4 m diameter has a surface charge density of 80.45 ¥ 10 -3 C fi 2 mg + 6qE = 4mg . The friction will act in opposite direction to the resultant of these forces. The driving forces on particle B are Fe and mg sin a.q ¯   b is maximum when sin q = 1 a   Example 17    Two charges +q1 and –q2 are placed at A and B b £ sin -1 ( m cot a ) ng respectively. m (m < tan a). If the particle is in equilibrium g 1 Q◊R df = dq ◊ E = (l ◊ 2 dy) ◊ in 4pe 0 ( R + R 2 )3/2 2 1 Q. The forces acting on B perpendicular to the line AB are component In of friction force f sin q and (mg sin a) sin b. what along the axis of the ring and then released.q ˆ Le =Á f = mg sin a ◊ sin b T Ë 4pe 0 2 2 R3 ◊ m ˜¯ sin q 1/2 f £ m mg cos a Ê 4pe 0 2 2 R3 ◊ m ˆ T = 2p Á ˜ Ë Q. Find the period T of td would be the maximum value of the angle b? the small amplitude oscillations of the rod.q a= ◊ dy rn 4pe 0 2 2 R3 ◊ m Ê 1 Q. Electrostatics: Part 1  65 . A line of force emanates from q1 at angle a with the line AB. For the particle at B to stay in equilibrium. di Let us analyses the forces parallel and perpendicular to line AB. At what angle will it terminate at –q2?   Example 16    In a free space. .indd 65 3/31/2017 4:32:10 PM . The rod is displaced slightly Module-Vol-II_01.L vt Solution If we displace the rod by length by upward the unbalanced force will be due to length 2dy. The ring is fixed and the rod is free to move. aP Solution Hence unbalance force on the element The electrostatic repulsive force will act along the line joining two charges along line AB.q ¯ mg sin a ◊ sin b ge £ m mg cos a 1/2 sin q Ê 2 2 pe 0 Rm ˆ = 4p R Á ˜ sin b £ m cot a ¥ sin q Ë Q. Let friction act at angle q from line AB. a thin rod carrying uniformly distributed negative charge –q is placed symmetrically along Ce the axis of a thin ring of radius R carrying uniformly distributed charge Q. Mass of the rod is m and length is l = 2R.q ˆ Hence w 2 = Á Ë 4pe 0 2 2 R3 ◊ m ˜¯ a f sin q = ( mg sin a ) sin b 1/2 2p Ê 1 Q. or + y=0 y0 dt 2 y0 aP at b is q2 2g ◊ 2p (1 . Í 2˙ . Æ q1 a q b ◊ 2 sin 2 = 2 ◊ 2 sin 2 a uniform electric field E = E iˆ is switched on and the ring starts In 0 2 2 2 2 rolling without sliding. Solution In equilibrium position. 4p y0 By the property of lines of force.L out equally in all directions. Solution Qq td It is the property of the lines of force that their number within . telec – fR = Ia Ce p /2 Ú l R dq E0 (2 R sin q ) .cos a ) or mg Í1 .cos b ) the charge per unit length l is as shown in the figure. At the base of the cylinder is a Solution point mass of charge Q having the same sign as q. which is constrained to move inside a narrow frictionless cylinder. 2 gy d 2 y 2g Similarly the number of lines of force terminating on –q2 or a=.   Example 19    A non-conducting ring of mass m and radius R.cos b ) which is equation for SHM with w = .indd 66 3/31/2017 4:32:12 PM . 66  Electrostatics: Part 1 . dF = lR dq E0 ng Net torque due to electric field t elec. q1 q di ◊ 2p (1 . from Newton’s second law. such that y < < y0. Find the angular frequency of the mass. gravitational force is balanced by ge Coulombic repulsive force Qq mg = 4pe 0 y02 a If charge q is displaced in positive y-direction.mg = ma 4pe 0 y02 ÎÍ (1 + y / y0 ) ˚˙ Hence lines of force per unit solid angle are q1/4p and the number q È 2y ˘ of lines through cone of half-angle a is 1 ◊ 2p (1 . it will exhibit Le simple harmonic motion. It is then 4p 4p placed on a rough non-conducting horizontal plane.mg = ma 4pe 0 ( y0 + y)2 a tube remains unchanged and the number of lines of force is equal to the charge. At time t = 0. Determine the friction force (magnitude b a q1 and direction) acting on the ring.fR = mR 2a 0 Module-Vol-II_01. its equilibrium position and released. sin = sin 2 2 q2 g Ê a q1 ˆ fi b = 2 sin -1 Á sin q2 ˜¯ in Ë 2   Example 18    A small point mass m has a charge rn q.˙ .mg = ma 4p Î y0 ˚ vt because solid angle of a cone is 2p (1 – cos a). = Ú dt Torque due to electrical charges dt = dF (2R sin q) Torque equation. The line of force emanating from q1 spreads Qq È 1 ˘ . Consider a differential element subtending an angle dq at the a If the mass m is displaced by a small amount from centre and at angle q as shown in figure.cos a ) = 2 ◊ 2p (1 . kx   Example 20    Two small identical balls lying on a horizontal dt 2 4pe 0 h 2 l 2 vt plane are connected by a weightless spring. kx Solution dt 2 Ë h ¯ h In 1 k When the balls are uncharged. When balls are charged the ball 1 will oscillate about new Comparing with a = –w2x g equilibrium position.indd 67 4/3/2017 12:13:15 PM .Á + k˜ x  …(ii) Ë 4pe 0 h l 3 3 ¯ From (i) and (ii) di d2x Ê 2(h .2 m = -Á k + k˜ x = .2 Thus the frequency is increased times h When the ball 1 is displaced by a small distance from the Here h = 2 and so frequency increases 2 times. Ê 1 2q 2 ˆ     = . Electrostatics: Part 1  67 ..k (hl + x . The balls are charged = ÁË1 .kl (h .1) . (ii) and (iii).1) . the unbalanced force to the right is given by Le a ge ng Ce Module-Vol-II_01.hl ˜¯ . -2 d2x 1 q2 Ê xˆ m = ÁË1 + hl ˜¯ . we get f = lRE0 along positive From Newton’s law. (3h . or 2l R 2 E0 .2 q2 = fi 3 l = h rn  . One ball (ball 2) is 1 q2 Ê 2x ˆ fixed at O and the other (ball 1) is free.2 ˆ k q2 w2 = fi v= 1 h 2p ÁË h ˜¯ m in = k (hl .1) ˆ 3h . we have x-axis.(i) v0 4pe 0h 2 (h ..2 ) k 1 Ê 3h . Determine the change in frequency. v0 = .fR = mR 2a  …(i) Resultant force to the right td and force equation f = ma …(ii) 1 q2 F= . x or a = .2 k Ê (3h .1) m 4pe 0 (hl )2 or v 3h . where k is force d2x 3h .l ) = kl (h .l ) and for rolling a = Ra …(iii) 4pe 0 (hl + x )2 .Á ◊x dt 2 h m Ë mh ˜¯ constant of the spring and m = mass of the oscillating ball (ball 1).L Solving equations (i).kx 4pe 0 h 2 l 2 identically as a result of which the spring length increases h = 2     aP times.1) k 3h . a equilibrium position to the right. The equilibrium position of ball 1.kl (h .2)k ˆ 2p m =. In figure. of radii r and R. and follows the trajectory x-axis are equidistant from the y-axis. Two small balls each having charge q are suspended by two insulating threads of equal length L from a hook in an elevator. with R > r. to their (f) What is the direction of the net force on the middle right. 2. ge      a (b) Does the Coulomb force that one charge exerts on ng other charges change if the other charges are brought nearby? (Yes/No) 8. greater than. the particles on the fired at the first particle. Module-Vol-II_01. why not? Give reason to support your answer. If these charges are kept at the same (c) Is the equilibrium stable or unstable? distance in water. Calculate the angle between two threads and tension in each thread. Suppose we have a large number of identical particles. What is the magnitude and direction of the net electrostatic force on the central particle due to the other particles? In 5. however. Is the angular momentum of experiences an electrostatics force F from each of these second particle constant about some axis? Why or two particles. Figure shows two charge particles on an axis. or less than 2F? particle can be placed such that all three particles are in (c) Does the x components of the two forces add or vt equilibrium. In each. How many particles were there in the group? a 7. the force between the charges is F´’. A second particle with positive charge q is Le +q or –q are fixed in place. 68  Electrostatics: Part 1 . g 6. a central particle of charge –q is surrounded by The ratio F´’/ F is equal to ……………. All three      Ce charges have the same magnitude. The charges (b) Is the magnitude of the net force on the particle on are free to move. Any two of them at 10 cm separation in repel with a force of 3 × 10–10 N. At one point. two circular rings of charged particles. (a) A particle with positive charge Q is held fixed at the 3..L different? 1. or between them? particle? (b) Should the third particle be positively or negatively 4. how strongly do you expect it to be rn repelled? (b) Suppose you measure the repulsion and find it 6 × 10–6 N. The elevator is freely falling.indd 68 3/31/2017 4:32:13 PM . Force between two point electric charges kept at a distance charged? di d apart in air is F. (a) If one of them is at 10 cm from a group (of very small size) of n others. Figure shows some of the electric field lines due to three point charges arranged along the vertical axis. Figure shows four situations in which particles of charge origin. NCERT Level Exercises td Subjective Type (a) Are the magnitude F of those forces the same or . The particle on y-axis as shown in the figure. a third charged y-axis equal to. cancel? (d) Do the y components of the forces add or cancel? (e) Is the direction of the net force on the middle particle aP that of the cancelling components or the adding components? (a) Is that point to the left of the first two particles. very small in size. find the angular frequency... if any. from the charge of 5 vt × 10–19 C. Two point charges of +5 × 10–19 C and +20 × 10–19 C are separated by a distance of 2 m. The electric field intensity will be zero at a distance d = . A in 10. 16.. g cm from each other... then the 9. What is the electric field strength? 12. Which of the following statement is correct? If E = 0. Sketch lines of force and locate the t proton neutral point.. For small oscillations about its equilibrium position.. An electron (mass me) falls through a distance ‘d’ in a aP uniform electric field of magnitude E... separated ng by a distance of 6 m.. td 13... Two charged metal plates in vacuum are 10 cm apart. Electrostatics: Part 1  69 .. flux is found to be f.. (c) the electric field for r > R2 is given by…………….. The electric field ge charge carried by the sphere is +Q. 20. A small electric dipole of dipole moment P is placed near Ce (a) the electric field for r < R1 is zero..indd 69 3/31/2017 4:32:14 PM . The total signs are placed a certain distance apart. 15.. An electron is released between the plates from rest at a point just outside the negative plate...L tetrahedron of which the triangle is the base. 14. It is placed in a uniform electric field of 8×104 V/m–1. Two point electric charges of unknown magnitudes and inner and outer radii of R1 and R2 respectively.. The (b) The total charge enclosed by the surface is zero.. What are the points at which resultant electric distance.. a point charge +Q as shown. rn Calculate: (i) How long (t) will electron take to reach the other plate? a (ii) At what velocity (v) will it be going just before it hits 11. until the position of maximum electric flux is found. (b) At what point(s) is the magnitude of the electric field the smallest? Explain your reasoning. Find the intensity of the field at the vertex of a regular . Two point charges Q and 4Q are fixed at a distance of 12 t ratio electron = .. at the other plate? Le all points of a closed surface: 17. this charge is uniformly intensity is zero at a point not between the charges but on distributed between R1 and R2. The side of the triangle is a. The direction of the field is reversed keeping its magnitude In (c) Two point charges q and –q are placed at a distance d unchanged and a proton (mass mp) falls through the same apart.. Is an electric field of the type shown by the electric lines uniform electric field of intensity (45/16) × 103 N/C–1 is in figure physically possible? applied between the plates. A hollow dielectric sphere as shown in the figure has 18. Then the net force on the (b) the electric field for R 1 < r < R 2 is given by dipole is towards …….. If the time taken by electron and proton to fall field is parallel to the line joining the two charges? the distance d is ‘telectron’ and ‘tproton’ respectively. respectively. (a) What are the signs of each of the three charges? Explain your reasoning. Explain how di the fields produced by each individual point charge combine to give a small net field at this point or points.. Write two essential conditions for this to happen... …………… Module-Vol-II_01.. Then the line joining them. A dipole consists of two particles carrying charges +2mC and −2mC and masses 1 kg and 2 kg. a 19. A ring of diameter d is rotated in a uniform electric field (a) The electric flux through the surface is zero. Three identical positive charges Q are arranged at the vertices of an equilateral triangle... Can two balls having same kind of charge on them attract 30. (iii) b < r < c. Force of attraction between two point electric charges to only those charges. 23. What are points at which the resultant field is parallel (iii) inner surface of the large shell. How this condition vt can be justified. S1 and S2 are two hollow concentric spheres enclosing td charges Q and 2Q respectively as shown in figure. apart. (c) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only. Define intensity of electric field at a point. What is the ratio of the electric flux through S1 and S2? . (c) What is the total charge on the in Or (i) inner surface of the small shell. 25. 70  Electrostatics: Part 1 . so that (b) Gauss law is applicable only when there is a force between them becomes F/3? symmetrical distribution of charge. Two point electric charges of unknown magnitude and 29. (v) r > d. the test charge has to be vanishingly small. By what factor will the (i) r < a. What distance apart should these be kept in the same medium. 28. a ge ng Ce Module-Vol-II_01.indd 70 3/31/2017 4:32:14 PM . rn to the line joining the two charges? (iv) outer surface of the large shell ? 26. line of an electric dipole of very small length? (b) Calculate the electric field (magnitude and direction) in terms of q and the distance r from the common 24. Write two essential conditions for this to inner shell has total charge +2q and the outer shell has happen. charge +4q. (ii) a < r < b. di on axial line and at a point at same distance on equatorial r………………. Which of the following statements is/are correct? each other? Explain. when we know that charge less than that on an electron or a proton is not possible? 22. The electric field intensity and outer radius b is concentric with a larger conducting is zero at a point not between the charges but on the line spherical shell with inner radius c and outer radius d. At what points is Show your results in a graph of the radial component g the electric dipole field intensity parallel to the line joining Æ the charges? of E as a function of r. electric field due to the dipole change? (iv) c < r < d. (a) Electric field calculated by Gauss law is the field due a 27. Two point charges +q and –q are placed at a distance d (ii) outer surface of the small shell. which are enclosed inside the Gaussian surface. What is the ratio of the strength of electric field at a point (a) Make a plot of the magnitude of the electric field vs. The joining them..L 21. A small conducting spherical shell with inner radius a aP sign are placed a distance apart. In defining electric field due to a point charge. The distance of the field point on the equatorial plane of In centre of the two shells for a small electric dipole is halved. Le placed at a distance d in a medium is F. Two charges q1 and q2 separated by a dielectric of dielectric constant 4 repel each other with a force of 10 N. Two identical metal balls with charges +2Q and −Q are (a) neutral (b) made of metal separated by same distance. (c) The interior becomes positively charged and the a (c) Both will acquire negative charge. A cube of side ‘b’ has charge ‘q’ at each of its vertices. A negatively charged cube ebonite rod is brought near A. 32q F F di (c) (d) Zero (a) become (b) become b 2 3 9 3. then the force of q1 exerted (d) proportional to its amplitude on q2 13. Now (a) The cube becomes negatively charged (a) A will have +ve charge and B will have –ve charge. (d) The interior remains charge free and the surface gets Le 5. The ratio of the forces acting on them will be (a) F (b) F/2 (a) 1 : 1 (b) 1 : 25 (c) F/4 (d) F/8 (c) 1 : 5 (d) 5 : 1 2. A positive point charge Q is brought near an isolated metal rn contact and insulated from earth. 1) show electrostatic 14. A simple pendulum of time period T is suspended above given an equal negative charge. F (c) become (d) remain F In are placed on the circumference of a circle of radius ‘r’ to 27 form an equilateral triangle. Two identical metallic spheres X and Y have exactly equal non-uniform charge distribution masses.indd 71 3/31/2017 4:32:15 PM . Pairs (1. ge (b) Mass of Y is greater than X its time period of oscillation will be (c) Mass is not involved (a) > T (d) Mass of X is greater than Y (b) < T 6. 5) show repulsion.L 1. (b) The cube becomes positively charged (b) A will have –ve charge and B will have +ve charge. Charge Q is divided into two parts which are then kept ng (a) will increase in magnitude some distanced apart. An electric charge q exerts a force F on a similar electric The electric field at the centre of the cube will be charge q separated by a distance r. 3). 2). (4. surface becomes negatively charged (d) Both will remain uncharged. If a (c) T a third charge q3 is brought near. 4) and (4. State Level Exercises td Single Correct Answer Type attraction. a large horizontal metal sheet with uniformly distributed (a) Masses of X and Y are equal positive charge. Two similar conducting spheres A and B are brought in 11. If the bob is given some negative charge. each carrying a positive charge Q. The force between them will now be vt coulomb. Therefore ball 1 must be . There are two charges +1 micro coulomb and 5 micro removed. which is then 8. aP 9. Then after charging. Five balls numbered (1 to 5 are suspended using separate (d) e and (Q-e). where e = electronic charge threads. A charge q1 exerts some force on a second charge q2. Another Module-Vol-II_01. Now. Electrostatics: Part 1  71 . Three small spheres. The electric field intensity at 10. the force (a) (b) b 2 2b 2 F will…. The force between them will be (b) will decrease in magnitude maximum if the two parts are (c) will remain unchanged (a) Q/2 each (d) will increase if q3 is of the same sign as q1 and will (b) Q/4 and 3Q/4 Ce decrease if q3 is of opposite sign (c) Q/3 and 2Q/3 7. A charged particle moves with a speed v in a circular path the centre of the circle will be of radius R around a long uniformly charged conductor 3Q 3Q 1 (a) v µ R (b) v µ g (a) (b) r r2 R Q 1 in (c) (d) zero (c) v µ (d) v is independent of R 2r 2 R 4. (2. They are joined by a conducting wire. X is given a positive charge q coulomb and Y is 12. A third charge q/4 is q q placed mid-way between the two charges. and exert a force F on each (c) positively charged (d) negatively charged other. while pairs (2. brought near to M. (a) −q (b) . If we place another charge Q at the zero. 1 All the three charges are positive. What should be A point charge q is placed at the centre of the shell and Q in order in make the net force on q to be zero? in another charge q1 is placed outside it as shown in the figure. charge q3 is placed between q1 and q2 such that the distance (a) the system can never be in equilibrium td of q3 from q1 is 1/4 times the distance of q3 from q2. –a] on the y-axis. If the resultant electric field on Q is equilateral triangle. Which diagram best illustrates the The charge Q will resultant distributions of charge on S and M ? (a) execute simple harmonic motion about the origin (a) (c) (b) (d) In (b) move to the origin and remain at rest (c) move to infinity (d) execute oscillatory but not simple harmonic motion 22. Q and q are placed in a straight line of g 17. (b) Q = . l/2 and l respectively. each corner is at a distance r from the centre. (b) the system will be in equilibrium if the charges rotate the force of repulsion between q1 and q2 is about the centre of the triangle (a) 10 N (b) 10 q1 (c) the system will be in equilibrium if the charges have . then the force on Q will be q a (a) Q = .5 (b) 0. forming an corners of the square. then centre of the circle. The force on the apart.1) k Le r2 r2 n q n -1 q (c) k 2 (d) k 2 (a) towards left (b) towards right n -1 r n r (c) upward (d) zero 24. An uncharged metal object M is insulated from its points [0. A charge Q is placed at each of two opposite corners of ge 18. –2l. length l at point 0.q 2 The force on the charge at the centre is (c) −2q (d) 4q rn 23. 72  Electrostatics: Part 1 . Three point charges are placed at the corners of an Ce equilateral triangle. is (in metre) (b) –ve direction of Y-axis (a) 1.5 (c) +ve direction of Z-axis (c) 1 (d) 2 (d) –ve direction of Z-axis 21. The field at the centre is a q q (a) k (b) (n . An B of charge –1 mC are placed in air at a distance 1 metre electric field acts on it in y-direction.L different magnitudes and different signs 4 q2 10 q1 (d) the system will be in equilibrium if the charges have (c) ¥ 10 N (d) q1 q2 the same magnitude but different signs Æ 20. A positive charge di surroundings. A charge q is laced at each of the two opposite on the circumference of a circle of radius R. A positively charged metal sphere S is then Q is released from rest at the point [2a. The electric field at the centre is 19. Identical charges of magnitude q are placed at (n – 1) corners. An electron moves with velocity v in x-direction. Assuming only electrostatic forces are acting: Module-Vol-II_01.2 2 q 1 qQ 2 2 (a) zero (b) ¥ 4pe 0 R 2 ng (c) Q = –2q (d) Q = 2 2 q 1 2qQ 1 3qQ 25. Two equal negative charges (–q each) are fixed at the 16. A point charge A of charge +4 mC and another point charge vt 15. In a regular polygon of n sides. Then the distance of the point on the line joining the electron acts in charges and from the charge B.indd 72 3/31/2017 4:32:16 PM . Three charges 4q. l and –l respectively. 0] on the x-axis. The charge per unit length of the four quadrant of the ring (c) ¥ 2 (d) ¥ 2 4pe 0 R 4pe 0 R is 2l. a] and [0. A thin metallic spherical shell contains a charge Q on it. where the resultant electric aP (a) +ve direction of Y-axis field is zero. Now. each carrying a charge q are placed a square. Three small spheres. (d) 180∞. A particle of charge – q and mass m moves in a circle of radius r around an infinitely long line of charge of linear 1 Q2 1 Q2 Ce charge density +l. 4pe 0 (2 L )2 4pe 0 L2 27. i (b) ˆj 2pe 0 R 2pe 0 R 1 2k l q aP 1 m (c) T = (d) T = 2l ˆ 2p r m 2p r 2k l q (c) i (d) None 4pe 0 R 1 where k = 26. passes through O and moves for 1. Two small spheres each having the charge +Q are suspended by insulating threads of length L from a hook. then the angle between the two a suspensions and the tension in each will be (d)  [KCET 2011] ng 1 Q 2 1 Q2 (a) 180∞. [EAMCET 2011] g R R (a) Q 2Q in (b) Q2 = .indd 73 3/31/2017 4:32:18 PM . 4pe 0 2 L2 4pe 0 L2 4. when (a) Q2 = Q Q . (b) 90∞. Q1 = Q . td . This arrangement is taken in space where there is no gravitational effect. An isolated solid metallic sphere is given +Q charge. The maximum its direction of motion.L vt m 4p 2 m (a) T = 2 p r (b) T 2 = l ˆ l 2 kl q 2k lq (a) . A negatively charged Archives di particle starts on the x-axis at a large distance from O. Electrostatics: Part 1  73 . Q1 = rn 4 4 Q Q (b) (d) Q1 = . Then time period will be: (c) 180∞. Which of the following best represents the plot? In  force of repulsion between them will occur. Q1 = Q - 4 3 Q 3Q (c) Q2 = . The charge will be distributed on the sphere[MH CET 2011] Le (a) Uniformly but only on surface (b) Only on surface but non-uniformly (c) (c) Uniformly inside the volume (d) Non-uniformly inside the volume ge 3. moves along the x-axis. At 4p Œ0 equal distance from the origin O. Two identical positive charges are fixed on the y-axis. If the force between q1 and q2 is F12 and that Module-Vol-II_01. A total charge Q is broken in two parts Q1 and Q1 and they away from O. Its acceleration a is taken as positive along are placed at a distance R from each other. The particle’s acceleration a is plotted against its x-coordinate. Three equal charges are placed on the three corners of a square. Q2 = 2 2 a 2. 19 20 (a) 10 (b) 10 Le  [KCET 2012] (c) 1.5 × 10–3 N (attractive) vt attraction.34 × 1042 atom Module-Vol-II_01. 1. Pear 3. Cylindrical They experience a force F. 8..L (a) 1/2 (b) 2 (a) 2. (d) Made of metal The magnitude of the net electric force on C is 6. The one which is best suited to retain the charges for a longer time (a) F (b) F / 2 is [KCET 2014] (c) F / 3 (d) 2F a (a) 1 (b) 2 10. Five balls numbered 1 to 5 are suspended using separate (c) 1. . The spheres are kept (a) Positively charged fixed with a distance ‘r’ between them.5 × 10–3 N (repulsive) threads. Four metal conductors having different shapes (c) Charged negatively (d) An insulator  [Kerala 2015] 9.4 × 10–3 N (attractive) (c) 1 / 2 (d) 2 (b) 2.6 C charge will be 8. will be (Charge on electron = 1. identical metal spheres A and B Therefore ball 1 must be [UP SEAT 2012] repel each other with a force ‘F’.. the ratio of magnitudes 12 is other by a force of 6 × 10–3 N. then the new force acting on each Ê 1 + 2 2 ˆ q2 Ê 1 ˆ q2 charge is  [MH CET 2014] (c) Á (d) Á 2 + ˜ Ë 2 ˜¯ 4pe 0 a 2 Ë 2 ¯ 4pe 0 a 2 (a) F / 8 (b) F / 4 g (c) 4 F (d) F / 16 7. A sphere 2. A third identical. induced charge in the aluminum will be[MP CET 2012] If a charge of 1 C is kept at the origin. 12. Two equally charged.4 × 10–9 N (attractive) 5. If each of them is given an F13 additional charge –6 × 106 C. 5) show repulsion. then what is the net (a) Zero force acting on 1 C charge? [EAMCET 2014] rn (b) Greater than in copper (a) 9000 N (b) 12000 N (c) Equal to that in copper (c) 24000 N (d) 36000 N (d) Less than in copper 15.. Pairs (1.6 × 10–27  [WB JEE 2015] kg. Lightning conductor additional charge of – 2mC. The ratio of electrostatic and gravitational forces acting (c) 3 (d) 4 ng between electron and proton separated by a distance 5 × 17.34 × 10 (d) 2. electrons from the earth a  [MPPET 2014] flow into the body. When a body is earth connected. mass law correctly describes the electric force that of electron = 9. 3) and (4. G = 6. mass of proton = 1. the new force of attraction will be [MH CET 2013] are mounted on insulating stands and charged. Two charges of equal magnitudes and at a distance r exert a (a) (b) force F on each other. Identify the wrong statement in the following. Two point charges 3 × 10–6 C and 8 × 10–6 C repel each td between q1 and q3 is F13. Two identical conductors of copper and aluminum are 14. aP (b) Negatively charged but uncharged sphere C is brought in contact with A and (c) Neutral then placed at the mid-point of the line joining A and B.C. 2. each of charge 1 mC..indd 74 3/31/2017 4:32:18 PM .1 × 10–31 kg. The number of electrons in 1. 4) and (4. 1) show electrostatic (d) 1.B. The magnitude of placed on the x-axis with co-ordinates x = 1. the force between them will  [MP PET 2011] be [KCET 2013] .6 × 10–19 C.1 × 102 (a) Unchanged (b) Charged positively 16. Coulomb’s 10–11m. F 11.•. The magnitude of the force on the (a) F (b) 3F/4 di charge at B will be [EAMCET 2012] (c) F/2 (d) F/4 3q 2 4q 2 13.1 × 1019 (d) 1.D of  [UPSEAT 2014] a square of length a. are in placed in an identical electric fields. (2.36 × 10 (b) 2. 74  Electrostatics: Part 1 . The charges on two sphere are +7mC and – 5mC respectively. 2). If each of them is given and ge 3.7 × 10–11 Nm2/kg2) [EAMCET 2013] (a) Binds the electrons of an atom to its nucleus Ce 39 (a) 2. 4. This means the body is…. If the charges are halved and distance In 4pe 0 a 2 4pe 0 a 2 between them is doubled. Equal charges q are placed at the four corners A.36 × 1040 41 (b) Binds the protons and neutrons in the nucleus of an (c) 2. while pair (2. An infinite number of charges. The total electric flux passing through the cylindrical surface is [MP PET 2016]  EAMCET 2015] 2 2 (a) 2pR E (b) pR /E (c) (pR2 – pR)/E (d) Zero vt 19. Electrostatics: Part 1  75 . A charge q is placed at the centre of the open end of a cylindrical vessel.L for the surface of the cylinder is given by wire as shown in the figure. q2. Electric charge is uniformly distributed along a long td (d) Binds atoms and molecules together to form solids straight wire of radius 1 mm. A point charge +q is placed at the centre of a cube of side shown in the figure and S is a spherical Gaussian surface L. Which of the following is true according to  [UPSEAT 2016] the Gauss’s law? [KCET 2016] q in (a) (b) Zero e0 6qL2 q rn (c) (d) e0 6 L2 e 0 22. q1. Eight dipoles of charges of magnitude e are placed inside a cube. The electric flux emerging from the cube is g of radius R. (c) Binds atoms together to form molecules 24.dA = e0 q 2q     (c) (d) (q1 + q2 + q3 + q4 ) 2e 0 e0 (c) Ú s (E1 + E2 + E3 ). The total flux radius 50 cm and length 1 m symmetrically encloses the . electric field of an infinitely (d) None of the above long straight wire is proportional to [MH CET 2016] 1 a (a) r (b) r2 ng 1 1 (c) 3 (d) r r Ce Module-Vol-II_01. q3 and q4 are point charges located at points as 21.indd 75 3/31/2017 4:32:20 PM . Another cylindrical surface of electric field E parallel to the cylinder axis. According to Gauss’ theorem. The flux of the electric field through     q1 + q2 + q3 the surface of the vessel is [EAMCET 2016] (a) Ú s (E1 + E2 + E3 ). A cylinder of radius R and length L is placed in a uniform of the wire is Q coulomb. The charge per cm length 18. Total electric flux coming out of a unit positive charge put in air is [KCET 2015] aP (a) e0 (b) e0–1 (c) (4pe0)–1 (d) 4pe0 20.dA = 2e 0 Le q     (q1 + q2 + q3 ) (a) Zero (b) e0 (b) Ú s (E1 + E2 + E3 ).dA = ge e0 23. The total electric flux coming out of the cube will be  [MP PET 2015] Q 100Q di 8e 16e (a) (b) (a) (b) e0 e0 e0 e0 10Q 100Q (c) (d) In e (pe 0 ) (pe 0 ) (c) (d) Zero e0 25. Another charge is kept in such a way (a) mV/qE (b) 2mV/qE that they do not move if released. field. Three charges of q1 = 1 × 10–6 C. q2 = 2 × 10–6 C and (d) E B = E A . 2r (a) 0 (b) 200pb2 2 (c) q. q2. Find the flux through the left end of cylinder. initially at rest. q3 and q4 is 1. r (d) 4q. q2. A by holding a positively charged rod very closely nearby. uniform electric field E has been applied in the vertical direction as shown. flux through surface S1 is nearly equal to (b) (mp/me)1/2 ng (a) 1 1/2 (c) (me/mp) (d) 1836 9. the ratio t2/t1 is a Then. The value of the third a (c) 3mV/2qE (d) 3mV/4qE charge and its position from –4q is 3.indd 76 3/31/2017 4:32:21 PM .2 E B (d) twice the total flux through S due to charges q3 and q4 if q1 + q2 = q3 + q4 di Æ Æ Æ 1 Æ 6. The figure shows four charges q1. and E B is perpendicular to E A 2 q3 = –3 × 10–6 C have been placed as shown. q4 fixed in space. 2r (c) 100pb (d) –200pb2 8. Charge on an originally uncharged conductor is separated a certain distance in a uniform electric field in time t1. also. A cylinder of length L and radius b had its axis coincident Le Æ with the x-axis. A and B are far away from the dipole and at equal distances from it. A and B are two points on the axis and the perpendicular bisector respectively of an electric dipole. due to all charges q1. moves through ge 4. In a gravity-free space. Assume that the induced negative charge on move through an equal distance in this uniform electric the conductor is equal to the positive charge q on the rod. JEE Exercises td Single Correct Answer Type 5. Then the total flux of electric field through a closed surface . A particle of mass m and charge q is projected horizontally with velocity V. Four point charges are placed at the corners of a square with diagonal 2a as shown. What is the total electric field Ce at the centre of the square? (a) zero (b) q/e0 (c) –q/e0 (d) None of these Module-Vol-II_01. proton of mass mp. initially at rest. The vt Æ Æ fields at A and B are E A and E B . Two charges q and –4q are held at a separation r on a frictionless surface. takes time t2 to as in figure. q3. is equal to in (a) S1 (b) S2 rn (c) S3 (d) same for all three 7. 2r (b) –4q. The time after which the velocity vector makes an angle g 45° with the initial velocity vector. 76  Electrostatics: Part 1 . (a) –2q. Neglecting the effect of gravity. An electron of mass me. Then the net electric flux will be maximum for the surface In 2. The electric field in this region is E = 200 iˆ . (a) not equal to the total flux through S due to charges q3 Æ Æ aP (a) E A = E B and q4 (b) equal to the total flux through S due to charges q3 and Æ Æ (b) E A = 2 E B q4 (c) zero if q1 + q2 = q3 + q4 Æ Æ (c) E A = .L Level 1 S. The block is made to oscillate in gravitation field. they hang in equilibrium as shown. Now two portions of the rn hemisphere are cut from either side and remaining portion is shown in figure. A block of mass m is suspended in vertical orientation an equilateral triangle. At how many points on the What would be the direction of the resulting electric field axis will the magnitude of electric field be E0/2 at the centre point P? (a) 1 (b) 2 di (c) 3 (d) 4 14. We now increase the charge on the left pith ball from Q to 2Q while leaving its mass essentially unchanged. Electrostatics: Part 1  77 . then electric field intensity at centre due to 3 remaining portion is E (a) 0 a 3 (a) (b) E0 ng (b) 6 E0 (c) 2 (c) (d) (d) Information insufficient Ce 12. Which of the following diagrams best represents the new equilibrium configuration? p ge If a = b = . Its time period is found to be Module-Vol-II_01. Now. The maximum electric field at a point on the axis of a cm square as shown. Three charges each Q are placed at the three corners of 16. q td centre of triangle. Four electrical charges are arranged on the corners of a 10 13. Two pith balls each with mass m are suspended from 15. a Le When the pith balls are given equal positive charge Q. The ratio so as to make the system in equilibrium is: Q . the force F will (a) become F 3 (b) become F 9 g (a) → (b) ↑ F (c) become (d) remain F in (c) ← (d) ↓ 27 11. charged hemispherical shell is E0.L (a) kq/a2 at an angle 45° above the +x-axis vt (b) kq/a2 at an angle 45° below the –x-axis (c) 3kq/a2 at an angle 45° above the –x-axis (a) 1: 3 (b) 1 : 3 (d) 3kq/a2 at an angle 45° below the +x-axis aP (c) 3 :1 (d) 2 : 3 10. A third charge q/4 is In placed mid-way between the two charges. An electric charge q exerts a force F on a similar electric charge q separated by a distance r. The electric field intensity at the centre of a uniformly insulating threads.indd 77 3/31/2017 4:32:22 PM . A fourth charge q is placed at the with a spring of spring constant k. uniformly charged ring is E0. We conclude Le 2 2 (b) q = (16/15) pe 0 mg L sin q tan q (c) q = (15/16) pe 0 mg L2 sin 2 q tan q (d) none of these 22. the foils separate even further. Two charges q1 and q2 are kept on x‑axis and electric field vt at different points an x‑axis is plotted against x. in (III) (IV) rn (a) all of the above (b) none of the above (c) II. Choose correct statement about nature and magnitude of q1 and q2. |q1| > |q2| shown can be correct? (b) q1 +ve. III and IV 18. q2 –ve . hang from three strings as shown in figure. aP qE (a) T (b) T + 2p md qE (c) 2p (d) none of the above md di 17. |q1| > |q2| In 21. L and q. q2 +ve . The new time 2 PkQ 2 PkQ period of oscillation is (c) . each of mass m and charge q. 78  Electrostatics: Part 1 . |q1| < |q2| Three identical point charges. q2 –ve . q2 +ve . ge Which one of the following diagrams correctly represents the electric lines of forces? (a) that the object is positively charged (b) that the object is electrically neutral a (c) that the object is negatively charged (d) none of these ng 19. An electroscope is given a positive charge. Which of the four situations (a) q1 +ve. III (d) II. A metallic shell has a point charge ‘q’ kept inside its cavity. When an object is brought near the top 2 2 (a) q = (16/5) pe 0 mg L sin q tan q a plate of the electroscope. causing its foil leaves to separate. cos q iˆ (b) cos q iˆ r r direction. Now the space between the plates is made gravity free PkQ PkQ td and an electric field E is produced in vertical downward (a) . cos q iˆ (d) cos q iˆ .L r 3 r3 Level 2 20. |q1| < |q2| (c) q1 –ve. A point negative charge – Q is placed at a distance r from (a) (b) a dipole with dipole moment P in x−y plane as shown in figure. The value g (I) (II) of q in terms of m. The x component of force acting on the charge – Q is Ce (c) (d) Module-Vol-II_01. T.indd 78 3/31/2017 4:32:24 PM . (d) q1 –ve. The figure below shows the forces that three charged particles exert on each other. Now the block is given charge q. A and is zero at the exact centre rod of length 2l has a linear charge density l on one half 29. Find the force experienced by the semicircular rod charged di induced charge density on right surface of the plate. The direction (q) of E at point P due to uniformly charged s l l2 td (a) 0 (b) sin q finite rod will be 2e0 s l l2 sll (c) sin q (d) . The rod is hinged at mid-point radius R. A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the centre. placed as shown in figure. An uncharged aluminium block has a cavity within it.indd 79 3/31/2017 4:32:25 PM . Find the 24. In g (a) – e0E (b) e0E (c) – 2e0E (d) 2e0E in 28. If the electric field ge O and makes an angle q with the normal to the sheet.  23. A large sheet carries uniform surface charge density s. An uncharged conducing large plate is placed as shown. It has volume charge density r. What is the ratio of the magnitude of vt the electric field at a distance 2R from the centre to the magnitude of the electric field at a distance of R/2 from the centre? aP (a) at angle 30° from x-axis (a) 1 (b) 2 (b) 45° from x-axis (c) 4 (d) 8 (c) 60° from x-axis 27. Which of the following is a rn correct statement describing conditions in the interior of the block’s cavity? lq lq (a) The electric field in the cavity is directed upwards (a) (b) a 2 (b) The electric field in the cavity is directed 2 p e0 R 2 p e0 R downwards Le lq lq (c) There is no electric field in the cavity (c) 2 (d) 4 p e0 R 4 p e0 R (d) The electric field in the cavity is of varying magnitude 25.L e0 2e0 26. Module-Vol-II_01. Consider an infinite line charge having uniform linear charge density and passing through the axis of a cylinder. Electrostatics: Part 1  79 . (d) none of these Now an electric field E towards right is applied. The at a point 2R distance above its centre is E then what is the torque experienced by the rod is electric field at the point which is 2R below its centre? a ng (a) rR/6e0 + E (b) ρR/12e0 – E Ce (c) –rR/6e0 + E (d) rR/12e0 + E 30. The block is placed in a region where a uniform electric field which is directed upwards. The diagram shows a uniformly charged hemisphere of and –l on the second half. Radius of the wire is R and the line of charge with linear charge density λ is passing through its centre and perpendicular to the plane of wire. with a charge q. ng field.6 p × 10–2 m2 (b) 0. The electric field E at 35. A positively charged sphere of radius r0 carries a volume charge density r (figure). Four very large metal plates are given the charges as shown any point P having position vector. A non-conducting sphere of radius R is filled with uniform (c) right (d) right rn volume charge density –r. Ë 4pe 0 ˜¯ aP An electric field E exists parallel to x−y plane. as shown. The sheet experiences a net electric force Ce of 0.L vt (a) Decreases (b) Increases (c) Remains same (d) Can’t say Ê 1 ˆ Use Á = 9 × 109 Nm2/C2 31. Find the area of one face of the sheet. Le ge r  r   (a) 5Q from A to B (b) 5Q/2 from A to B (a) d (b) (r .9 p × 10–2 m2 flux through remaining curved part. Find the (a) 3. The middle two are then connected through a a wire. A charged large metal sheet is placed into uniform electric the cone is h. One fourth of a sphere of radius R is removed as shown. C1 is the di centre of sphere and C2 that of cavity. A conic surface is placed in a uniform electric field E as r   r  shown such that field is perpendicular to the surface on the (c) (d . Assume external field to remain constant after introducing the large sheet. What is the direction and magnitude of the electric field at point B? In g 17 rr0 rr0 (a) p R 2 E (b) 2p R 2 E (a) left (b) left in 54 Œ0 6 Œ0 (c) p R 2 E / 2 (d) None of these 17 rr0 r r0 32. What will be the effect on the flux passing through curved td surface if the portions of the line charge outside the cylinder is removed? . The centre of this sphere is 54 Œ0 6 Œ0   displaced from the origin by d .8 p × 10 m (d) none 34. 80  Electrostatics: Part 1 .d ) (c) 5Q from B to A (d) no charge will flow 3 Œ0 3 Œ0 36. Find the charge that will flow through the wire. After placing the sheet into the field. inside the sphere is in figure.indd 80 3/31/2017 4:32:26 PM . the electric field on the left side of the sheet is E1 = 5 × 105 V/m and on the right it is E2 = 3 × 105 V/m. Module-Vol-II_01. The base of the cone is of radius R and height of 33. –2 2 (c) 1. perpendicularly to the electric field lines.r ) (d) (r ) a 3 Œ0 3 Œ0 side AB. A spherical cavity of radius r0/2 is then scooped out and left empty.08 N. –y). B and C respectively then: di (a) q/e0 (b) q/2e0 (c) q/4e0 (d) zero 38. 2e 0 (a) ER[h cos q + p(R/2) sinq] . Consider a sphere of radius R with O as centre and R > y. The angle of cone is q as shown. Three large identical conducting parallel plates carrying aP charge +Q. Two point charges (Q each) are placed at (0. y) and view is shown in the figure. along x-axis.indd 81 3/31/2017 4:32:28 PM . Then: Four points are marked I. E 3 and E 4 (b) the charge q is in equilibrium at the origin (c) the charge q performs an oscillatory motion about the ng respectively. Find the magnitude of td   s 12 + s 22  that flux which enters the cone curved surface from left (a) E1 = E 2 = π E4 side. III and IV. A point charge q of the same polarity can move uniformly on sheets in electrostatic equilibrium condition. Don’t count the outgoing flux. (q < 45°). The number of electric field lines crossing an area DS g   is is n1 when D S  E . II. Electrostatics: Part 1  81 . -Q and +2Q respectively are placed as shown in the figure. while number of field lines crossing   same area is n2 when D E makes an angle of 30º with E. Flux passing through shaded surface of sphere when a point     s 12 + s 22 (c) E1 = E 2 = E 3 = E 4 = vt charge q is placed at the centre is (radius of the sphere is 2e 0 R) (d) None of the above 40. When an electron moves in a circular path around a 2l R (a) zero (b) stationary nucleus charge at the centre: e0 Le (a) the acceleration of the electron changes (b) the velocity of the electron changes 2l R2 . Two infinite sheets having charge densities s1and s2 are ge placed in two perpendicular planes whose two-dimensional 2. The correct expression for electric field intensities is origin (d) for any position of q other than origin the force is directed away from origin 3. Electric flux through the surface of the sphere In (a) EA > EB > EC (c) EA = 0 and EB > EC (b) EA = EB >EC (d) EA = 0 and EB = EC 41. E 2. in then: (a) n1 = n2 (b) n1 > n2 (c) n1 < n2 (d) cannot say anything rn Multiple Correct Answers Type a 1. EB and EC refer the magnitude of electric field at points A.y2 l R2 + y2 (c) (d) (c) electric field due to the nucleus at the electron e0 e0 changes (d) none of these 39.L (b) ER[h sinq + p(R/2) cosq]   s 12 + s 22 (b) E 2 = E 4 = (c) ER[h cos q + pR sinq] 2e 0 (d) none of these 37. A conducting ball is positively charged and another positive Ce point charge is brought closer to the ball. If EA. The electric (a) the force on q is maximum at x = +y/ 2 a     field intensities at these points are E1 . The charges are distributed (0. A uniformly charged and infinitely long line having a linear charge density ‘l’is placed at a normal distance y from a point O. (a) the ball may attract the point charge (b) the ball may repel the point charge Module-Vol-II_01. L (d) Field at point S is Á E + equal and opposite point charges at the ends of an insulating Ë A Œ0 ˜¯ rod of length d. then (d) B (5 cm from the charge) is 1. covering a dipole (c) B (5 cm from the charge) is 0 in of charge q and –q.5 × 10–6 C at the centre (c) will feel a torque trying to make it rotate of a spherical cavity of radius 3 cm of piece of metal. A simple pendulum of mass m charged negatively to q Ce coulomb oscillates with a time period T in a downward electric field E such that mg > qE. Each dipole consists of two Ê Q ˆ . both E and f will change (b) If Q2 changes. (a) If Q1 changes. 8. The dipole on the left: vt f be the flux of E over S. then (c) E = 0 at all points on the spherical surface Le  (d) E · ds = 0 (Surface integral of over the spherical Ú surface) 7. If points (a) f A = fC π 0 (b) f D π 0 R. C and D. We have two electric dipoles. B. the new time period (b) field at point S is E Module-Vol-II_01. (c) there may be no force between them Ê Q ˆ td (d) the ball will only repel the point charge and in no (c) field at point T is Á E + Ë Œ0 A ˜¯ condition it can attract the point charge 4. Consider a Gaussian spherical surface. fB. Area of each plate is A and charges +Q and –Q are given to these plates as shown in the figure. B and C are the points in the plane containing rn the base of the cone. then which of the following cannot be the magnitude of electric field anywhere on the surface of sphere (a) 4 N/C (b) 8 N/C (a) A (2 cm from the charge) is 0 g (c 16 N/C (d) 32 N/C (b) A (2 cm from the charge) is 1. If the electric field is (a) field at point R is E withdrawn. oriented as shown below.indd 82 3/31/2017 4:32:29 PM . while D is the point at the vertex of (a) q in = 0 (Net charge enclosed by the spherical the cone. Let E be the field at any point on S and Their separation r >> d.8 × 106 N/C 10. fC and fD represent the flux through surface) curved surface of the cone when a point charge Q is at a (b) fnet = 0 (Net flux coming out the spherical surface) point A. then the Q (c) f B = (d) f A = f C = f D = 0 2e 0 a ng Linked Comprehension Type For Problems (1–2) 1. Two large thin conducting plates with small gap in between are placed in a uniform electric field ‘E’ (perpendicular to ge the plates). S and T as shown in the figure are three points in space. respectively. A right circular imaginary cone is shown in the adjacent figure.125 × 107 N/C 6. 82  Electrostatics: Part 1 . E will change but f will not change aP (c) If Q1 = 0 and Q2 ≠ 0 then E ≠ 0 but f = 0 (a) will feel a force to the left (d) If Q1 ≠ 0 and Q2 = 0 then E = 0 but f ≠ 0 (b) will feel a force to the right 9. A. If magnitude of electric field at a certain point on the In surface of sphere is 10 N/C. The figure shows a point charge of 0. If fA. The counterclockwise electric field at (d) will feel no torque di 5. Charges Q1 and Q2 lie inside and outside respectively of a closed surface S. Imagine a short dipole is at the centre of a spherical surface. The dipoles sit along the x‑axis a distance r apart. Then we gently charge densities r and –r. For Problems (6–7) td Positive and negative charges of equal magnitude lie along the symmetry axis of a cylinder.indd 83 3/31/2017 4:32:30 PM . Determine the maximum g The spheres are now laid down such that they overlap as shown velocity attained by the ball which is closer to the axis in Æ Æ the subsequent motion. At equilibrium of the bob the change in tension in string cylinder? will be (assuming rest condition) (a) 0 (b) + Q / e0 (a) mg (b) qE (c) + 2Q / e0 (d) – Q / e0 (c) 2qE (d) qE/2 7. The distance from the positive charge to the left end-cap of the cylinder is the same as the .d a gives electric flux Module-Vol-II_01. Take d = O1O2 in 2QEL 2QEL (a) (b) m 5m rn QEL 4QEL (c) (d) 5m 5m 4. (a) = T (b) > T vt (c) < T (d) any of the above three is possible 6. What is the sign of the flux through the right end-cap of For Problems (3–5) the cylinder? di There is an insulator rod of length L and of negligible mass with (a) Positive two small balls of mass m and electric charge Q attached to its (b) Negative ends. In what position is the rod to be set so that if displaced a a little from that position it begins a harmonic oscillation  about the axis A? 8. in the figure. For Problems (8–9) 3.L distance from the negative charge to the right end-cap. The electric field E in the overlap region is Le (a) non-uniform (b) zero r Æ -r Æ (c) d (d) d 3 Œ0 3 Œ0 9. Both spheres have equal radius R. Gauss’s law is e0f 5mL 5mL = qencl in which qencl is the net charge inside an imaginary closed (c) 2p (d) 2p   8QE 4QE surface called Gaussian surface f =  Ú E. Electrostatics: Part 1  83 . The rod can rotate in the horizontal plane around a vertical (c) There is no flux through the right end-cap. The potential difference DV between the centres of the two ge (a) (b) spheres for d = R is r 2 r 2 (a) d (b) d 3 Œ0 Œ0 a 2r 2 (c) (d) (c) zero (d) d Œ0 ng 5. What is the flux of the electric field through the closed aP 2. displace it from this position. In axis crossing it at an L/4 distance from one of its ends. At first the rod is in unstable equilibrium in a horizontal There are two non-conducting spheres having uniform volume uniform electric field of field strength E. What is the time period of the SHM as mentioned in above question? For Problems (10–13) mL 2 mL Gauss’s law and coulomb’s law expressed in different forms (a) 2p (b) 2p although are equivalent ways of describing the relation between QE 3QE Ce charge and electric field in static conditions. indd 84 3/31/2017 4:32:32 PM . A Gaussian surface encloses two of the 4 positively charged 15. Find the tensions in the string BC. of through Gaussian surface. B. +2q and +4q in remain same. Find the tensions in the string AB. (a) Both A and B are true (b) Both A and B are false a (c) A is true but B is false 17. The particles which contribute to the electric field inside the cavity) then the correct representation of field . 84  Electrostatics: Part 1 . The net flux of the electric field through the surface due q3 (q1 + q2 ) to q3 and q4 is (a) zero (b) 4pe 0 r 2 (a) zero (c) negative (b) positive (d) can’t say 13. then consider the following In (c) -q3 (q1 + q2 ) 4pe 0 r 2 (d) q3 q1 + q2 q3 + q1q2 4pe 0 r 2 g two statements For Problems (17–20) A: Electric field at each point on Gaussian surface will Three particles A. are connected by strings as shown in the figure. (c) (d) none of these 4pe 0 d 2 there is a point charge q1 at centre of one cavity and q2 at the centre of other cavity. ge distance ‘r’ from the centre of the spherical conductor.L at point p on the surface are lines inside the same cavity is: vt (a) (b) (a) q1 and q2 (b) q2 and q3 (c) Then will be no field lines inside cavity aP (c) q4 and q3 (d) q1. If the particle C is discharged then tension in string BC 3 q2 (a) zero (b) 14. However. 3 q2 (a) zero (b) 4pe 0 d 2 9 q2 (c) (d) none of these a 4pe 0 d 2 ng 19. If q1 is displaced from its centre slightly (being always particles. The two equations hold only when (c) Charge q3 applies equal force on both the charges td the net charge is in vacuum or air. Another charge q3 is placed at large 18. The net flux of the electric field through the surface is (a) due to q1 and q2 only (b) due to q3 and q4 only (c) equal due to all the four charges (d) di (d) cannot say 16. The total charge on conductor itself is zero. If the charge q3 and q4 are displaced (always remaining outside the Gaussian surface). The force acting on conductor A will be 12. (d) B is true but A is false 3 q2 (a) zero (b) Le For Problems (14–16) 4pe 0 d 2 A spherical conductor A contains two spherical cavities as shown 9 q2 in figure. q3 and q4 11. (d) Charge q3 applies no force on any of the charges 10. d A for the Gaussian surface will E rn remain same. Assuming   equilibrium of the particle. B: The value of  Ú . q2. Which of the following statements are true? 4pe 0 d 2 Ce (a) Charge q3 applies larger force on charge q2 than on 9 q2 (c) (d) none of these charge q1 4pe 0 d 2 (b) Charge q3 applies smaller force on charge q2 than on charge q1 Module-Vol-II_01. C having charges +q. S3 and S4. Surface of plates are numbered (1). Column I Column II (p) Magnitude of flux through (a) p R2 E In base of cone (q) Magnitude of flux through (b) curved part of cone p R2 E 2 g (r) Magnitude of flux through (c) zero curved part MNQP of in Column I Column II cone (p) Surfaces having charges of (a) 1 and 8 (s) Net flux through entire (d) non-zero same magnitude and sign cone rn (q) Surfaces having positive (b) 3 and 5 charges 4. Inside a neutral metallic spherical shell a charge Q1 is Le placed and outside the shell a charge Q2 is placed ge Column I Column II Column I Column II (p) If Q1 is at the centre (a) Electric field inside (p) Net flux through S1 (a) Q of shell the shell remains zero - a e0 (q) If Q1 is not at the (b) Electric field inside (q) Net flux through S2 (b) ng centre of shell the shell remains 2Q non-zero - e0 (r) If position of Q1 is (c) Electric field inside changed within the changes (r) Net flux through S3 (c) zero shell keeping Q2 (s) Net flux through S4 (d) non-zero Ce fixed Module-Vol-II_01. (2). 4 p e0 d 2 vt (d) is equal to the tension in BC. The charges on each plate are given in diagram and separation between plates is d (d is very small). Four large parallel identical conducting plates of area A are arranged as shown. It is kept in a uniform electric field E parallel 7 q (c) to its axis. 20. Electrostatics: Part 1  85 . +Q. –Q and four closed (r) Uncharged surfaces (c) 2 and 3 surfaces S1. S2.L (b) 4pe 0 d 2 3. (s) (d) a Charged surfaces 6 and 7 2. ……… di (8). Matching Column Type aP 1. Axis of a hollow cone shown in figure is vertical. Its base 2 radius is R. If the particle B is discharged then the tension in string (s) If position of Q2 (d) Electric field outside td AB is changed outside changes (a) becomes zero keeping Q1 fixed inside at any point 5 q2 .indd 85 3/31/2017 4:32:33 PM . Consider three point charges –2Q. Two identical small equally charged conducting balls are suspended from long threads secured at one point. In the figure shown a conducting spherical shell of inner O radius x and outer radius y is concentric with a larger conducting spherical shell of inner radius a and outer radius A B a b. The inner shell has a total charge +3Q and the outer shell has a total charge +5Q. q. C and D) of a square pyramid (iv) fS3 (s) 2Q with slant length ‘a’ (AP = BP = BP = PC = a = 2 m ). strength is the larger spherical non-zero shell (iii) Magnitude of (r) On outer surface of . r (b) s p. r s r p (ii) fS2 (q) 0 Integer Answer Type In (iii) fS3 (r) Q - e0 1. A dipole with dipole moment p = 1 C-m is placed at centre of base and perpendicular to in Codes its plane as shown in figure. Calculate the charge (q) that should be placed at II. ng 3. the centre of the ring such that the electric field becomes zero at a point on the axis of the ring distant ‘R’ from the Q centre of the ring. charge particles is N . S3 and (ii) Electric field (q) On inner surface of td S4. Column I Column II (a) p q s p. Find the value of W. iv. r p (i) fS1 (p) Q + (c) q.L charge is 3Q the smaller spherical shell (iv) Charge is + 8Q (s) For a < r < b vt Now match the given columns and select the correct option from the codes given below. 5. q. Four charge particles each having charge Q = 1 C are fixed at corners of base (at A. If the force on dipole due to Ω i. Let r be the distance of any point 2. A ring of radius R. 4pe 0 rn (a) s r q p P (b) q s r p (c) q r s p D C a a 2 (d) r s p q a Le 6. q. r p p. i. If the value of q = Ω . Codes aP Now. b Module-Vol-II_01. iii. iv. iii. ii. S2. Match column I and column over it. strength is zero larger spherical shell Again for the certain value of distance b (b << l) between a3 the balls the equilibrium is restored. The charges and masses of the balls are such that they are Column I Column II in equilibrium when the distance between them is a (the Ce (i) Electric field (p) On outer surface of the length of thread L >> a). Find the value of 3 . ii. One of the balls is then discharged.indd 86 3/31/2017 4:32:34 PM . match the following columns. a - g e0 charge -Q is fixed at point P. has charge –Q distributed uniformly ge from the common centre O. 86  Electrostatics: Part 1 . q. r s e0 di (d) p. r q. B. Following figure shows for Gaussian surface S1. Find the value a 4 of W. Consider a uniform charge distribution with charge thread. Taking g = p2 and length of the simple pendulum 1 m. nF. Let their approach velocity v varies as v µ x -1/2 . If this particle is moving with a are suspended from some point on the ceiling of a damp uniform horizontal speed of 0. 4. Due to ionization of medium. 11 aP di  6. Find the value of k. S1 and S2 are two hollow concentric spheres with charge q (in sec). +Q on +q charge is F. . charge leaks off from density 2 C/m3 throughout in space. Find the value of N. find the time period of the simple pendulum 8. a distance between sphere be x. The surrounding medium is air with coefficient of viscosity rn 7. (r = radius of sphere). Electric field in a region is given by E = . Find the value of n. Then. The In charge enclosed in the cube of side 1 m oriented as shown in the diagram is given by a Œ0 .L is 1 mg. then net electric force on the +Q is 9K find the value of . N ¥ 69 in 10. If a Gaussian sphere Le each sphere and they keep on coming closer to each other has a variable radius which changes at the rate of 2 m/s.28 × 105 V/M. Two small metal sphere having equal charge and mass h = 1. at a constant rate.4 xiˆ + 6 yjˆ. Speed of 1 g electron in Bohr’s orbit is about ¥ speed of light . the number of excess room with silk threads of equal length.02 m/s. a ge ng Ce Module-Vol-II_01. The charge of the bob is 1 mC and mass Find the value of N. A simple pendulum is suspended in a lift which is going If mass of each sphere is m then the rate at which charge td up with an acceleration of 5m/s2. x << l. Find the value of a. an electron moves in a circular orbit about a proton of radius 0. Six charges are kept at the vertices of a regular hexagon vt as shown in the figure.indd 87 3/31/2017 4:32:34 PM . dt 2 l present in the lift. In Bohr’s theory of the hydrogen atom. then value of rate of charge of flux is proportional to rk. Then. 5. If magnitude of force applied by The ratio of flux through S2 and flux through S1 is K.6 × 10–5 N-s/m2. An electric field of dq N 2pe 0 mg magnitude 5 N/C and directed vertically upward is also varies with respect to time is µ . Electrostatics: Part 1  87 . and 2q.53 Å. A charged dust particle of radius 5 × 10–7 m is moving in a horizontal electric field of intensity 6. find the value of k. Let centre to centre electrons on the drop are 6k. Space between S1 and S2 is filled with a dielectric of dielectric constants. 9. l is length of silk 11. If a charge q is placed at the centre of the line joining two spherical conductor. If it is allowed to oscillate above a positively charged metal plate.(1 + 2 2 ) (d) (1 + 2 2 ) 4. Another same charge q is placed vt X at a distance L from O. JEE Archives td JEE (Main) � q3 Y . are f1 and f2. + q2. Four charges equal to – Q are placed at the four corners of a a 4 4 square and a charge q is placed at its centre.indd 88 3/31/2017 4:32:36 PM . If the system is 3. Then the electric flux through � q1 � q2 ABCD is E F q2 q3 q2 q3 aP (a) . A charged particle q is placed at the centre O of a cube of b length L (ABCDEFGH). having the same radius as that of B equal charges Q such that the system is in equilibrium. The dipole will experience 5. cos q (b) + sin q b2 a2 b2 a2 D C q2 q3 q2 q3 (c) + cos q (d) . A third in 2. if the side of equilateral triangle is l cm? 9. and – q3 are placed as shown (a) a translational force only in the direction of the field Ce in the figure. The new force of repulsion between B and C is rn Q -Q (a) (b) (a) F/4 (b) 3F/4 2 2 (c) F/8 (d) 3F/8 (AIEEE 2004) Q -Q (c) (d)  (AIEEE 2002) 8. If three charges are placed at the vertices of equilateral 2 2 ge triangle of charge ‘q’ each. (1 + 2 2 ) (b) (1 + 2 2 ) (a) remain equal to T (b) less than T 4 4 (c) greater than T (d) infinite (AIEEE 2002) Q Q (c) . Two point charges + 8q and –2q are located at x = 0 and 1 q2 1 2q 2 = L. A simple pendulum of period T has a metal bob which is in equilibrium. The location of a point on the x-axis at (a) (b) which the net electric field due to these two point charges 4pe 0 l 4pe 0 l a is zero is 1 3q 2 1 4q 2 (a) 2 L (b) L/4 ng (c) (d) (c) 8 L (d) 4 L (AIEEE 2005) 4pe 0 l 4pe 0 l 10. Three charges – q1. is brought in contact with B and then in then the value of q is contact with C and finally removed away from both. The x-component of the force on – q1 is (b) a translational force only in a direction normal to the proportional to direction of the field (c) a torque as well as translational force (d) a torque only (AIEEE 2006) Module-Vol-II_01. sin q O b2 a2 b2 a2 q q  (AIEEE 2003) H G di 6.L Single Correct Answer Type q a 1. 88  Electrostatics: Part 1 . the electric charge inside the A B surface will be L (a) (f2 − f1)e0 (b) (f1 + f2)/e0 In q (c) (f2 − f1)/e0 (d) (f1 + f2)e0 (a) (b) zero 4pe 0 L  (AIEEE 2003) 7. Two spherical conductors B and C having equal radii q q g (c) (d)  (AIEEE 2002) and carrying equal charges on them repel each other 2pe 0 L 3pe 0 L with a force F when kept apart at some distance. respectively. but uncharged. If the electric flux entering and leaving an enclosed surface. An electric dipole is placed at an angle of 30° to a non-  (AIEEE 2002) uniform electric field. What is the net potential  (AIEEE 2004) energy. the value of q is Le negatively charged. its period will Q Q (a) . respectively. and r(r) = 0 for r > R. Let there be a spherically symmetric charge distribution . the angle remains the same. If the density of the 12. (b) - 3e 0 ÁË 3 R ˜¯ 4e 0 ÁË 3 R ˜¯ aP r r O R O R E(r) 4 r0 r Ê 5 r ˆ r0 r Ê 5 r ˆ E(r) (c) . Two identically charged spheres are suspended by strings of equal lengths. The electric field at a distance r (r < R) from the (a) (b) origin is given by 4pr0 r Ê 5 r ˆ r0 r Ê 5 r ˆ (a) . The strings make an angle of 30° with r each other. If the net electrical force on Q is zero. A thin spherical shell of radius R has charge Q spread q q td uniformly over its surface. The graph which would correspond to the (a) -2 2 (b) –1 above will be 1 (c) 1 (d) . the magnitude of electric field is point by two massless strings of length l are initially a distance d(d << l) apart becuase of their mutual repulsion. where r is the distance from vt the origin. Two identical charged spheres suspended from a common in sphere.indd 89 3/31/2017 4:32:38 PM .˜ up to E(r) r Ë 4 R¯ r = R. Electrostatics: Part 1  89 . Which of the following graphs (c) . A thin semi-circular ring of radius r has a positive charge (a)   (b) q distributed uniformly over it. 11. (d) - 3e 0 ÁË 4 R ˜¯ 3e 0 ÁË 4 R ˜¯  (AIEEE 2010) di (c) (d) 16. where r is the distance 15. As a result the charges approach each other rn Qr12 Qr12 with a velocity v.L from the center of the shell? Ê5 rˆ  (AIEEE 2008) with charge density varying as r (r ) = r0 Á . 2 2 j (d) j 2p e 0 r 2p e 0 r 2 2 most closely represents the electric field E(r) produced by  (AIEEE 2010) the shell in the range 0 £ r < •. the electric field E is plotted as function of distance Q/q equals from the centre. A charge Q is placed at each of the opposite corners of (c) v µ x  (d) v µ x (AIEEE 2011) a square. the dielectric constant of the liquid is (a) 4 (c) 2 (b) 3 (d) 1 (AIEEE 2010) g a solid sphere of radius R and total charge Q. Then as a function of distance x between (c) (d)  (AIEEE 2009) them 4pe 0 R 4 3pe 0 R 4 (a) v µ x –1/2 (b) v µ x –1 a 1/2 13. In a uniformly charged sphere of total charge Q and radius Le corners. A charge q is placed at each of the other two 18. then the R.  (AIEEE 2009) E E 2 ge 14. Q (a) 0 (b) The charge begins to leak from both the spheres at a 4pe 0 r12 constant rate. The net field E at the centre O is R r R r a j E E ng (c)   (d)  i O q q R r R r (a) (b) - Ce 2 2 j 2 2 j  (AIEEE 2012) 4p e 0 r 4p e 0 r Module-Vol-II_01. Let P(r ) = p R4 Q r be the charge density distribution for In material of the sphere is 16 g/cm3. When suspended in a liquid of density 0.8 O R r O R g/cm3. For a point p inside the sphere at distance r1 from the centre of the 17. A positive charge Q is released from rest at Statement 2: The electric field at a distance r (r < R) from the point (2a. each equal to q. The system of the three charges will be  (AIEEE 2012) in equilibrium if q is equal to 20. where A is a constant and r is the distance a r 5. The value of A such that the electric field in the force should be sketched as in (IIT-JEE 2001) region between the spheres will be constant. As a result of this uniform charge (c) (d)  (JEE Main 2016) π (a 2 . statement 2 is 2. Two equal negative charges –q are fixed at points (0. Let [e0] denote the dimensional formula of the permittivity a of vacuum. statement 2 is true. statement 2 is true. are kept at x = –a and (a) –Q/2 (b) –Q/4 x = a on the x-axis. At the centre of the spheres is a point charge Q. If charge q0 is given a small  (IIT-JEE 1987) displacement (y << a) along the y-axis.a 2 ) that best describe the two statements. is Ce (a) (b) (c) (d) Module-Vol-II_01. (d) execute oscillatory but not simple harmonic motion (c) Statement 1 is false statement 2 is true.L positive charge density r.  (IIT-JEE 1984) (d) Statement 1 is true. The resulting lines of ng from the centre. choose the one (a) (b) 2π a 2 2π (b2 . Two charges. then: (c) 3 (d) 4 (IIT-JEE 1996) Le (a) [e 0 ] = [ M -1 L-3T 4 A2 ] 4. the ratio t2/t1 is (d) [e 0 ] = [ M -1 L-3T 2 A]  (JEE Main 2013) ge nearly equal to 22. A charge Q is placed at the center of the line joining two In the correct explanation of statement 1. 90  Electrostatics: Part 1 . A metallic solid sphere is placed in a uniform electric in on the particle is proportional to 1 field. A particle of mass m and charge (c) +Q/4 (d) –Q/2 g q0 = q/2 is placed at the origin. L = length. Three positive charges of equal value q are placed at the vertices of an equilateral triangle.indd 90 3/31/2017 4:32:40 PM . 19. (d) y (JEE Main 2013) y 21. T = time and A = (a) 1 (b) 2 electric current. (c) move to infinity di (b) Statement 1 is true statement 2 is false. at the surface of the sphere and also at a point out side the sphere. This questions has Statement 1 and Statement 2. Of the Q Q td four choices given after the statements. An insulating solid sphere of radius R has a unioformly 2Q 2Q . The region between two concentric spheres of radii ‘a’ (a) 1 (b) (mp /me)1/2 and ‘b’. ± a) aP qr/3e0 on the y-axis.b2 ) π a2 distribution there is a finite value of electric potential at the centre of the sphere. 0) on the x-axis. equal charges Q. The lines of force follow the path(s) shown in the (a) –y (b) figure as y rn 1 (c) . If M = mass. has volume charge (c) (me /mp)1/2 (d) 1836 A  (IIT-JEE 1997) density ρ = . The charge Q will the centre of the sphere is rr/3e0 (a) execute simple harmonic motion about the origin (a) Statement 1 is true. Neglecting the effect of gravity. respectively (see figure). A proton of mass mp also initially at rest takes time t2 to (c) [e 0 ] = [ M -1 L2T 1 A] move through an equal distance in this uniform electric field. An electron of mass me initially at rest moves through a -1 2 1 -2 (b) [e 0 ] = [ M L T A ] certain distance in a uniform electric field in time t1. The electric potential at Jee (Advanced) vt infinite is zero. Statement 1: When a charge ‘q’ is take from the centre of Single Correct Answer Type the surface of the sphere its potential energy changes by 1. statement 2 is (b) move to the origin and remain at rest not the correct explanation of statement 1. the net force acting 3. and Q3. +. +. 6. +. 0) g The electric field on the surface will be and (−3a / 4. respectively. A metallic shell has a point charge q kept inside its cavity. A spherical portion has been removed from a solid sphere . ge It is found that the surface charge densities on the outer surfaces of the shells are equal. (c) negative and distributed nonuniformly over the entire td Which one of the following diagrams correctly represents surface of the sphere the electric lines of force? (d) zero (IIT-JEE 2007) 11. +. +. y = ± a / 2. respectively. +.L having a charge distributed uniformly in its volume as shown in the figure. −. − (b) −. q2 � q1 z = ± a / 2. +. 0) . −. such that field at the center is double that of what it would have been if only one positive charge is placed at R. three positive and three negative. 0) . − 6 C is placed in the xy plane with its center at (− a /2. −.indd 91 3/31/2017 4:32:42 PM . 2R. −. Then. Three infinitely long charge sheets are placed as shown in the figure. +. (a) zero everywhere P Q (b) nonzero and uniform (c) nonuniform di U R O (d) zero only at its center (IIT-JEE 2007) T S 12. The net charge on the sphere is then Ce (a) negative and distributed uniformly over the surface of the sphere (b) negative and appears only at the point on the sphere closest to the point charge Module-Vol-II_01. Consider a cubical in q1 surface formed by six surfaces x = ± a / 2. Q2. (c) −. The electric flux through this cubical surface is rn (a) due to q1 and q2 only (b) due to q2 only a (c) zero (d) due to all (IIT-JEE 2004) Le 9. Electrostatics: Part 1  91 . The electric field inside the empty (a) (b) (c) (d) space is vt  (IIT-JEE 2003) 7. +. − A rod of length carrying a uniformly distributed charge 8  (IIT-JEE 2004) C is placed on the x-axis from x = a / 4 to x = 5a / 4. A Gaussian surface in the figure is shown by dotted line. Two 8. and 3R are given charges Q1. point charges –7 C and 3 C are placed at (a / 4. Three concentric metallic spherical shells of radii R. A positive point charge is placed outside the sphere. the ratio Q1 : Q2 : Q3 of the charges given to the shells is a (a) 2s / e 0 kˆ (b) 4s / e 0 kˆ (c) -2s / e kˆ (d) -4s / e 0 kˆ  (IIT-JEE 2005) ng 0 10. are to be placed on PQRSTU corners of a regular aP hexagon. Six charges of equal magnitude. +. − a / 4. The electric field at point P is (a) –2C/e0 (b) 2C/e0 (c) 10C/e0 (d) 12C/e0  (IIT-JEE 2009) 13. −. −. A disk of radius a /4 having a uniformly distributed charge In (a) +. Consider a neutral conducting sphere. 0. − (d) +.3a / 4. (c) Charge +q executes simple harmonic motion while charge –q continues moving in the direction of its ge (a) 2E0 a 2 (b) 2E0 a 2 displacement. the magnitude resulting electric field. 0) (. When the field is switched off.2 ¥ 10–19 C to move in the x-direction only. E0 a 2 (d) Charge –q executes simple harmonic motion while (c) E0 a 2 (d)  (IIT-JEE 2011) 2 charge +q continues moving in the direction of its displacement. Given g = 9. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1. 2 and 3 are E1.8 ms–2.6 ¥ 10–19 C (b) 3. Charges Q. then the SHM of the block will be cavity of radius R2. then aP 1 2 2 1 2 (a) s R (b) s R e0 e0 1 s2 1 s2 (c) (d) e0 R e 0 R2 di  (IIT-JEE 2010) (a) E1 > E2 > E3 (b) E3 > E1 > E2 15.8 ¥ 10–19 C (d) 8. 0. 92  Electrostatics: Part 1 . 2 and 3 of radii R/2. and the density of oil is 900 kg m–3.0. centred at P with distance OP = a = R1 – R2 (see figure) is made. The point charges are confined (a) 1.a.L carries uniform surface charge density of s per unit area. 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1. (d) of changed frequency and with the same mean position It is made of two hemispherical shells. A uniformly charged thin spherical shell of radius R (c) of changed frequency and with shifted mean position . F is proportional to 18. a. In this distribution. In their Ns m–2. point charges q and –q are kept in g of q is equilibrium between them. then the correct statement(s) +Q is(are) Ce Module-Vol-II_01.indd 92 3/31/2017 4:32:44 PM . a) Le (a) Both charges execute simple harmonic motion. held together by  (IIT-JEE 2011) pressing them with force F. viscosity of the air is 1. Consider a uniform spherical charge distribution of radius ng surface. 0. then the  (IIT-JEE 2010) correct statement(s) is (are): → rn 16. If they are given a small in (c) 4.a. as shown in figure. Consider an electric field E = E0 xˆ . where E0 is a constant.0 ¥ 10–19 C displacement about their equilibrium positions. a) (. the 19. 0) their displacement.8 ¥ 10–5 charge density l are kept parallel to each other. R and 2R vt F F respectively. a. If a uniform electric field E is switched on as R1 centred at the origin O. (JEE Advanced 2015) a 17. (b) Both charges will continue moving in the direction of (0. If the electric field inside the    E cavity at position r is E (r ). The flux through the shaded area (as shown in the figure) due to this field is a (. The figure below depict two situations in which two drop is observed to fall with terminal velocity 2 ¥ 10–3 infinitely long static line charges of constant positive line ms–1. (a) 1:2:3 (b) 1:3:5 (a) of the same frequency and with shifted mean position td (c) 1:4:9 (d) 1:8:18 (b) of the same frequency and with the same mean  (IIT-JEE 2009) position 14. A tiny spherical oil drop carrying a net charge q is balanced (c) E2 > E1 > E3 (d) E3 > E2 > E1 in still air with a vertical uniform electric field of strength (JEE Advanced 2014) In  (81p/7) ¥ 105 Vm–1. E2 and E3 respectively. The block carries a charge +Q on its 20. a spherical shown. A wooden block performs SHM on a frictionless surface with frequency n0. A negatively In is more than the net electric flux crossing the plane charged particle P is released from rest at the point y = –a/2. 0. 0). for 0 < r < ∞ wire with constant linear charge density l. the center of the spheres is zero. A few electric field lines for a system of two charges Q1 (a) Q = 4sp r02 (b) r0 = and Q2 fixed at two different points on the x-axis are shown 2ps in the figure. + . for all values of z0 satisfying 0 < z0 < ∞ is equal to the net electric flux crossing the plane g (b) simple harmonic. for all values of z0 satisfying x = +a/2. The electric field is (c) at a finite distance to the left of Q1.  (a) E is uniform. 0). –q at (0. along the line joining rn the negative z-axis toward z = −∞. its magnitude depends of R2 and its  direction depends on r aP  (c) E is uniform. the electric field td is zero (IIT-JEE 2010) 4. 0). z0 ). (IIT-JEE 2012) 5. Then the motion of P is (c) The net electric flux crossing the entire origin is q/e0. If E1(r0) (d) is discontinuous at r = R (IIT-JEE 1998) = E2 (r0 ) = E3 (r0 ) at a given distance r0. its magnitude is independent of a but its  direction depends on a  (d) E is uniform. 0. (d) at a finite distance to the right of Q2. +a/4. Choose the correct option(s). Electrostatics: Part 1  93 . where z0 > 0. The charge density r(r) (charge per unit (a) |Q1| > |Q2| volume) is dependent only on the radial distance r from the Ce (b) |Q1| < |Q2| center of the nucleus as shown in the figure. the electric field is only along the radial direction. and –q at (0.indd 93 3/31/2017 4:32:47 PM . then l ge 3. Two nonconducting solid spheres of radii R and 2R. Let E1(r). A positively charged thin metal ring of radius R is fixed in (b) The net electric flux crossing the plane y = +a/2 the xy plane with its center at the origin O. The ratio r1 / r2 can be  (IIT-JEE 1998) (a) –4 (b) –32/25 2. (c) approximately simple harmonic. and an infinite (c) decreases as r increases. These lines suggest that Ê r0 ˆ Ê r0 ˆ Ê r0 ˆ Ê r0 ˆ (c) E1 Á ˜ = 2 E2 Á ˜ (d) E2 Á ˜ = 4 E3 Á ˜ Ë 2¯ Ë 2¯ Ë 2¯ Ë 2¯ a  (JEE Advanced 2014) Q Q ng 2 1 Linked Comprehension Type For Problems (1_3) The nuclear charge (Ze) is nonuniformly distributed within a nucleus of radius R. –a/4. It encloses three fixed point charges. (IIT-JEE 2008) zero Module-Vol-II_01. A cubical region of side a has its center at the origin. A nonconducting solid sphere of radius R is uniformly (c) 32/25 (d) 4 a charged. provided z0 << R touch each other. (0. respectively. The magnitude of the electric field due to the  (JEE Advanced 2013) sphere at a distance r from its center 6. E2(r) and E3(r) be the respective electric fields Le (a) increases as r increases. The net electric field at a distance 2R (d) such that P crosses O and continues to move along from the center of the smaller sphere. an infinitely long (b) decreases as r increases. for R < r < ∞ plane with uniform surface charge density s. 1. and both its magnitude and direction  depends on a  (JEE Advanced 2015) (a) The net electric flux crossing the plane x = +a/2 di is equal to the net electric flux crossing the plane Multiple Correct Answers Type x = –a/2.L 3q at (0. (d) The net electric flux crossing the plane z = +a/2 (a) periodic. for r < R at a distance r from a point charge Q. having in 0 < z0 ≤ R uniform volume charge densities r1 and r2. its magnitude is independent of R2 but vt  its direction depends on r  (b) E is uniform. Q-2. A positive charge q is placed on the positive 2 y axis at a distance b > 0. (IIT-JEE 2009) (d) inversely proportional to a aP 2. Q4 positive. –a. The direction of the forces on ABCD lying in the x–y plane with its centre at the origin is a the charge q is given in List II. as shown in 3Ze 3Ze the figure. The electric field at r = R is its volume with a charge density r = kra where k and a (a) independent of a are constants and r is the distance from its centre. For a = 0. Q4 negative (2) –x (r) Q1.The value of k is di (c) (d) 3p R 3 3p R 3 3. If the vt (b) directly proportional to a electric field at r = R / 2 is 1/8 times that at r = R. R-4. A solid sphere of radius r has a charge Q distributed in 1. find (c) directly proportional to a2 the value of a. 94  Electrostatics: Part 1 . Match List I with List II lL and select the correct answer using the code given below (e0 = permittivity of free space). R-3. 3 y-z plane at z = rn a (see figure). An infinitely long solid cylinder of radius R has a uniform figure) is volume charge density r. R-1. Q3 negative (3) +y ng (s) Q1. Q4 all positive (1) +x a (q) Q1. This implies R (a) a = 0 (b) a = 2 2R (c) a = R (d) a = g 3  (IIT-JEE 2012) in Matching Column Type 3. S-2 Ce (b) P-4. Q3. Q2.L Integer Answer Type 1. Q3 and Q4 of same magnitude charge per unit length l lies parallel to the y-axis in the are fixed along the x axis at x = –2a. which is at a distance 2R from the axis of the cylinder. The magnitude of the electric field at the point (a) (b) 4p R 3 p R3 P. The electric field within the nucleus is generally observed In to be linearly dependent on r. then the value of n ne 0 the lists. Q-1. Q2. Q3 positive. S-3  (JEE Advanced 2014) . Q-2. Le is __________ ge Column I Column II (p) Q1. An infinitely long uniform line charge distribution of 1. Q-1. Q4 negative (4) –y . Q2 positive. Q3. (c) P-3. Q2. the value of d (maximum value of r as shown in 2. Four charges Q1. It has a spherical cavity of radius R/2 with its center on the axis of the cylinder. 4 Ze Ze is given by the expression 23r R /16k e0 . If the magnitude of respectively. Q2.indd 94 3/31/2017 4:32:49 PM . S-4 td (d) P-4. (JEE Advanced 2015) Codes (a) P-3. R-2. Four options of the signs of these the flux of the electric field though the rectangular surface charges are given in List I. S-1 Module-Vol-II_01. +a and +2a. (d) 3. (ii). 3. –1. ng 2 Œ0 components. (a) 3.. (c) 13. e 0 E ..L 1. 0. x2 = a Qb ËÁ x1 ˜¯ 2 Ê  + x1 ˆ 3 ÁË x ˜¯ . (b) 5. (b) no directions of displacement.00 9... (b) (b) Less than 2F 6. (c) 8. (a) positive. Answers td Concept Application Exercise 1 NCERT Level Exercises . (c) 10.00 mC of the charge particles from its equilibrium position... (a) Right. (b) –25 14. (d) 3. because if we displace any 8. Charge b is negative and charge a is positive..2 2 .28 × 1013 C 11.00 mC. 11 × 10– 4 N-m2/C 16. 1.8 ¥ 10 4 (-2iˆ + 3 ˆj . (a) 4..indd 95 3/31/2017 4:32:50 PM . (d) 11. (a) 2. (b) 2. (a) 4. 29 29 rn (i) (ii) 14. −2.4 kˆ ) 13. (c) 6. a (e) Obviously direction of net force will be that of adding 15. (d) 2.. di 10 q2 12. (c) (iii) and (iv) 14. (c) (c) Cancel in situations (i) and (ii) and add in situations 11. (d) 7. (iii) and (iv) will be same. 0. (b) in 1.0 × 10−10 C (c) Equilibrium is unstable. but not between the vt 10.0. (c.. = ( 3 .375 7. (a) 9.19 MC 6. (f) (i) +y (ii) –y (iii) +x (iv) –x Ce Module-Vol-II_01. Concept Application Exercise 3 g 1. (a) Q / q = .25 10.. (a) 5. 0... cm (b) 0 2. (a) −1. (d) 3. (d) 12. only then Concept Application Exercise 2 net force on any charge due to other two charges can aP 1. Electrostatics: Part 1  95 .3 days charges. so on (d) Add in situations (i) and (ii) and cancel in situations a (iii) and (iv). (b) electron 8.1) 2pe 0 r 2 In 13. (b) Third particle should be negatively charged. +16e 1. because third particle should be placed near 9. (a) 9. (b) be zero.. (d) 2. (b) 9. 1. d) 4. (b) 1. (b) 1.3 × 107 C 7. (c) 2.0 The direction is at angle 45° from x-axis in IV-quadrant. 4 × 10–6 N (iii) (iv) Concept Application Exercise 4 (a) It is clear from diagrams that the forces F (F1 or F2) in cases (i).33 × 1010 electrons the charge smaller in magnitude. 2 Ê ˆ  Qa = 1 +  . (a) . (b) 3. (b) +9.. -e 0 E.1 Le 1 15. 6. (c) 5. (a) 4.50 C Subjective Type 5. (b) 12. (c) 8. it may not come back to its initial position for all 11. (d) 3. (c) 10.6 × 10−7 kg. (c) 7.00 × 1010 electrons. −4. (a) positron. ge 1. (a) 2 × 10–8 s 17. b. (b) 20. (b) 18. (b. (d) 15. (a. 180°. (a) 22. (a) 24. JEE Exercises 3 mp 16. (a) 24. (a) 12.2 rad/s 20. (a) 20. (c) 7. The resulting plot is as shown in figure. 35. (a) True (b) True Archives aP Q È r 3 . (c) (b) No 21. (d) 10. middle (–). (d) 12. electric field is zero. (a.L 1. d) q (iii) . (c. d) 7. (d) 4. 3 rn Level 2 29. d) 5. 5. (c) 33. Along U 1. (a) 3. (d) 18. (d) 2. (a) 20. (d) 2. (a) 7. (d) a (v) . 96  Electrostatics: Part 1 . (a) 15. (a) 14. d) 6. (b) 26. (c) 39. 6. (a) 10. d) 8. (a) n × 10 N  (b) 2 × 10 . (d) 24.R1 ˚˙ 4pe 0 r 2 6. b. (c) 26. (c) 21. (b) 25. electric field will be zero. (a) 5. (d) 19. as no charge 40. (d) 41. (c) 8. (c) 2. (b) 10. (a) (c) On perpendicular bisector of line joining the charges. 2 6. (a) False (b) False (c) True Ce Module-Vol-II_01. (b) 1. (d) 12. (a) 34. (c) 38. (b) Le inside. No 11. (a. (c) 29. radially outward 6. (d) 14. (b) 18. (a) 11. (b) 7. (b) ng (c) (i) 0 (ii) +2q (iii) –2q (iv) +6q 16. 25. 9. b. c) 10. 8 times 25. b. (a) 8. (b) (ii) for c > r > b. (a) (i) for r > b. Theory based 23. (d) 14. d) 3. (a) 31. (a) The angular momentum of q will be constant about Q. (c) 19. radially outward (iv) 0 Linked Comprehension Type 2 p e0 r 2 3q 1. because no charge 20. (b) 28. (c) 2. (a) Yes. (a) 4. (b) 9. (b) 27. (b) 13. x = 3 d 28. 2 2 pe 0 a 2 di 21. (d) 13. c) (b) (i) 0 (ii) 0 ge 9. (a) 37. 2 81 16pe 0 L Single Correct Answer Type –10 4 6. (b) inside. (c) 4. (b) 15. (c) 23. (c) 23. (a) 17. Theory based 27. (c) 1 16.indd 96 3/31/2017 4:32:51 PM . (b) 13. (b) and this force produces no torque about Q. (a) 8. (a) 22. (a. (c) 18. (b) 5. (d) 15. (a) 10. (b) 2 me 14. (b) 9. (a) vt (d) 8. Theory based 11. (a) 7. 16. (b) 25. (a) 3. (b. (a. X = m 15. (b) 3. electric field will be similar to that as if 30.R13 ˘ 1 Q 12. (b) in 24. (c) 2 Í 3 3˙ (c) 4pe 0 r ÎÍ R2 . (a) Top (+). c) 2. 4 cm 10. (a. (a) 36. (b) 21. (iv) for r > d. (a) a a point charge 2q is placed at centre. (c) 9. (b) 5. (d) 3. electric field will be similar to that as if a Multiple Correct Answers Type point charge 6q is placed at centre. Theory based 22. Theory based In Single Correct Answer Type Level 1 g 19. (a) 13. (c) 7. (a) 23. (b) 8. (a) 17. c) 4 (a. (a) 3Q 16. (d) 14. (d) 30. 0. (d) because force acting on q due to Q will pass through Q 11. bottom (+) (b) Two 26. (c) (iii) for d > r > c. 22. (d) 17. (a) 9. (a) 19. p d2 4f (b) 107 m/s 18. (d) 12. (a) 19. (b) 17. q2 State Level Exercises td 1 4. (a) 13. (d) 2 p e0 r 2 11. (b) 27. (a) 5. (c) 32. 1. (d) 4. (b. c) 6. (2) 10. d) 1. b. (6) 2. (a) 6. Matching Column Type JEE (Advanced) td 1. c) Single Correct Answer Type 2. d) 5. (d) 2. (c) . (a) 10. d) (s – c) 6. (5) aP 11. (a. 16. (c) 5. (a) 13. c) 2. (b) 3. (a) 18. (c) 15. (9) 9. d) (r – b. (a) 2. (p – a. b. (2) Linked Comprehension Type 1. (b) 20. Electrostatics: Part 1  97 . (c) (c) (a) 12. (c) 19. (b) 3. c. (c) 20. (3) 8. (d) 4. (c) 17. (d) 9. (c) 10. (2) 7. (a. (d) 3. (c) 19. (9) 6. (d) 5. (c) 7. (b) 5. d) (q – c) (r – b. c) (s – b. (b) 4. 21. (a) 15. (d) Integer Answer Type Multiple Correct Answers Type vt 1. (a. (d) 4. (c) 14. (b) 1. (c) 11. d) 1. (a) 22. c) 3. (b) 14. (a) 18. (p – b) (q – b) (r – b. d) (q – a. (d) In g in a rn Le a ge ng Ce Module-Vol-II_01. (2) 2. (a. (6) 3. (a) 13. (a) di Single Correct Answer Type Integer Answer Type 1. (4) 4.L 3. b) (r – d) (s – a. (a) 7. (b) 16. d) (s – c) 11. (a. (2) 3. (2) 5. (p – a) (q – a. (6) 6. d) 4. (b) 2. (c) 8. (c) 17. (p – a.indd 97 3/31/2017 4:32:51 PM . (b) 9. (d) 8. (b) 12. (c) Jee Archives Matching Column Type JEE (Main) 1. L vt aP di In g in rna Le a ge ng Ce Module-Vol-II_01. . td .indd 98 3/31/2017 4:32:51 PM .