TEACHING NOTESELECTROSTATICS-1 JEE Syllabus : Coulomb’s law; Electric field and potential; Electrical potential energy of a system of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; Gauss’s law and its application in simple cases, such as, to find field due to infinitely long straight wire, uniformly charged infinite plane sheet and uniformly charged thin spherical shell. Complete in 12 LECTURE Introduction A number of simple experiments demonstrate the existence of electric forces and charges. For example, after running a comb through your hair on a dry day, you will find that the comb attracts bits of paper. The attractive force is often strong enough to suspend the paper. Another simple experiment is to rub an inflated balloon with wool. The balloon then adheres to a wall, often for hours. When materials behave in this way, they are said to be electrified or to have become electrically charged. You can easily electrify your body by vigorously rubbing your shoes on a wool rug. Evidence of the electric charge on your body can be detected by lightly touching (and startling) a friend. Under the right conditions, you will see a spark when you touch, and both of you will feel a slight tingle. (Experiments such as these work best on a dry day because an excessive amount of moisture in the air can cause any charge you build up to “leak” from your body to the Earth. Electric and magnetic phenomenon are generally bracketed together, since both derive from charged particles. Magnetism, arises from charges in motion. However, in the frame of reference where all charges are at rest, the forces are purely electrical. The subject of electrostatics, as the name suggest deals with the physics of charges at rest. Electro (Related to charge) + statics (stationary). Hence it deals with stationary charges Properties of Electric Charges (i) Charge comes in two varieties, which are called “plus” and “minus”, Like charges repel each other and unlike charges attract each other. Only two kinds of electric charges exist because any unknown charge that is found experimentally to be attracted to a positive charge is also repelled by a negative charge. No one has ever observed a charged object that is repelled by both a positive and a negative charge. (ii) Charge is conserved : The charge of an isolated system is conserved. The algebraic sum of charges in any electrically isolated system does not change. (iii) Charge is quantized : Protons and electron are considered the only charge carriers. All desirable charges must be integral multiples of e. If an object contains n1 protons and n2 electrons, the net charge on the object is n1 (e) + n2 (– e) = (n1 – n2) e. Thus, the charge on any object is always an integral multiple of e and can be charged only in steps of e, i.e. charge is quantized. The step size e is usually, so small that we can easily neglect the quantization. Now, 1C contains n units of basic charge e where n= 1C ~ 6 × 1012 1.6 10 19 C The step size is thus very small as compared to the charges usually found on many cases we can assume a continuous charge variation. This was verified by millikan oil drop experiment. Page-1 Meaning of a charged body A material is said to be charged if there are more of one kind of charge than the other. A negatively charged body has excess electrons over protons, while a positively charged body has excess positive charges over electrons. The protons are tightly bound in the nucleus, making them very difficult to remove. Charging a body therefore involves the removal, addition, and rearrangement of the orbital electrons. A body becomes positively charged if it loses electrons, and negatively charged if electrons are added to it. Now we study ways in which this addition, removal or rearrangement is achieved in practice. WAYS OF CHARGING 1. Charging by friction Charging by friction is the oldest form of charging. It was found that when an amber rod is rubbed with fur, the rod became negatively charged. The two bodies acquire opposite signs of electricity ; one gets positively charged, while the other becomes negatively charged. When two bodies are charged by friction, they acquire the same magnitude of charge. Furthermore, the bodies retain these excess charges even when they are separated from each other. Note : Charging involves transformation of mass. 2. Charging by conduction In charging by contact, an uncharged body is brought into contact with a charged body. The uncharged body acquires the same sign of charge as the charged body. The total charge is distributed between the two bodies. uncharged body A 3. + + + + + + + + B + + + + + + + bodies retain charge on separation charging by contact charged body + + + + + + + + A + + B + + + + + + + A + + + + + + + + + B + + + + Charging by induction In charging by induction, an uncharged body is brought close to, but not touching, a charged body. Charging by induction is an example of the rearrangement of charges between bodies. The sign of the induced charge is opposite to that of the inducing charge. Furthermore, the induced charges last only while the inducing charge is present. The induced charged disappear when the inducing charge is taken away. (Charge can be retained if use grounding) Asking question Three objects are brought close to each other, two at a time. When objects A and B are brought together, they attract. When objects B and C are brought together, they repel. From this, we conclude that (a) objects A and C possess charge of the same sign. (b) objects A and C possess charges of opposite sign. (c) all three of the objects possess charges of the same sign (d) one of the objects is neutral (e) we need to perform additional experiment to determine information about the charges on the objects. Note : Neutral does not mean “chargeless”. Note : The bodies around is are almost neutral because there is microscopic balance of -ve and +ve charge. Page-2 Asking question If a negatively charged body is attracting a body of unknown nature of charged body What are the possible signs of the unknown charged body. – –– – – – – – – ? -ve charged body Answer It can have any charge as a –ve charged body can definitely attract +ve, charged body, It can as well attract neutral or -ve charged body because of induction positive charge on a neutral body will be closer to the -ve charge. Although the charge on +ve body should be small so that effect of the distance overrule the slight effect of repulsion. Note : Attraction is not the proof the nature of charge on bodies. COULOMB’S LAW From experimental observations on the electric force, Coulomb’s law can be expressed, giving the magnitude of the electric force of interaction between two point charges: Fe = ke | q1 || q 2 | r2 where ke is a constant The Coulomb constant ke in SI units has the value ke = 8.987 5 × 109 N·m2/C2 9 × 109 N·m2/C2 This constant is also written in the form 1 ke = 4 0 Ex.1 Sol. where the constant 0 (lowercase Greek epsilon) is known as the permittivity of free space and has the value. 0 = 8.854 2 × 10–12 C2 / N · m2 The electric force is a conservative force. The electron and proton of a hydrogen atom are separated (on the average) by a distance of approximately 5.3 × 10–11 m. Find the magnitudes of the electric force and the gravitational force between the two particles. Charge and mass of the electron and proton Particle Charge (C) Mass (kg) –19 Electron (e) – 1.602 1917 × 10 9.109 5 × 10–31 –19 Proton (P) + 1.602 191 × 10 1.672 61 × 10–27 From Coulomb’s law, (1.60 1019 C) 2 | e | | e | 9 2 2 Fe = ke = (8.99 × 10 N · m / C ) = 8.2 × 10–8 N (5.3 10 11 m) 2 r2 Using Newton’s law of universal gravitation Fg = G me m p = (6.67 × 10–11 r2 The ratio Fe / Fg 2 × 1039. N· m2 / kg2) × (9.11 10 31 kg ) (1.67 10 27 kg) = 3.6 × 10–47 N (5.3 10 11 m) 2 Gravitational force between charged atomic particles is negligible when compared with the electric force. Page-3 Coulomb’s law in vector form Remember that force is a vector quantity and must be treated accordingly. In vector form the force on charge q1 due to charge q2 is expresed as q1q 2 rˆ r2 F12 = ke {q1 & q2 should be substituted with sign.} ˆr is a unit vector directed from q2 toward q1 F12 r – + q1 + F21 F12 ^r q2 + F21 (b) (a) When the charges are of the same sign, the force is repulsive. When the charges are of opposite signs, the force is attractive. The electric force obeys Newton’s third law, the electric force exerted by q2 and q1 is equal in magnitude to the force exerted by q1 on q2 and in the opposite direction ; that is, F21 = – F12. If q1 and q2 are of opposite sign, as shown in figure (b), the product q1q2 is negative. A negative product indicates an attractive force, so that the charges each experience a force toward the other. Coulomb forces follows principle of Superposition. i.e. the resultant force on any one of the charge is equal to the vector sum of the forces exerted by the various individual charges. For example, if four charges are present, then the resultant force exerted by charge 2, 3 and 4 on charge 1 is F1 F12 F13 F14 Charges in medium : Consider two point charges q1 and q2 kept in a medium of permittivity r (r 1 and is 1 for vacuum). The medium will be consisting of atoms which are neutral but not chargeless. If any substance is present in vicinity of a charge, the positive and negative charges of atom (nuclei and electrons) experience electric force, which in turn leads to a partial separation of these charges. These partially separated charges present on atoms apply additional electric force on charges which in combination with the force of interaction between q1 and q2 gives the resultant field. If we know the force of interaction and the additional electric force due to induced charges, we can forget about the presence of the substance itself while calculating the resultant force, since the role of the substance has already been taken into account with the help of induced charges. Thus, the resultant force in the presence of a substance is determined simply as the superposition of the external field and the field of induced charges. r + q•1 – q•2 (medium is ) F1 = F12 + F force on 1 due to polarisation. 1 q1 q 2 F1 represents net force on 1 and = 4 r2 r 0 q1 q 2 But force on q1 due to q2 is still F12 = 4 r 2 0 1/ 2 Page-4 Eg. Very simple Eg Consider three point charges located at the corners of a right triangle as shown in figure, q1 = q3 = 5.0 C, q2 = = – 2.0C, and a = 0.10m. Find the resultant force exerted on q3. y q2 – a F13 F23 a + q3 2a x q1 + Sol. F23 = 9.0 N in the coordinate system shown in figure, the attractive force F23 is to the left (in the negative x direction). The magnitude of the force F13 exerted by q1 on q3 is F3 = 11N the repulsive force F23 makes an angle of 45° with the x axis. Combing F13 with F23 by the rules of vector addition. To demostrate the principle of superposition. Eg. Two identical small charged spheres, each having a mass of 3.0 × 10–2 kg, hang in equilibrium as shown in figure. The length of each string is 0.15m, and the angle is 5.0°. Find the magnitude of the charge on each sphere. //////////////////////////////// L L q a q L=0.15m = 5.0° Sol. Fx = T sin – Fe = 0 Fy = T cos – mg = 0 |q| = 4.4 × 10–8C There is no way we could find the sign of the charge from the information given. In fact, the sign of the charge is not important. The situation will be exactly the same whether both spheres are positively charged or negatively charged. (1) (2) Asking question Suppose we propose solving this problem without the assumption that the charges are of equal magnitude. We claim that the symmetry of the problem is destroyed if the charges are not equal, so that the strings would make two different angles with the vertical, and the problem would be much more complicated. How would you respond? Ans. You should argue that the symmetry is not destroyed and the angles remain the same. Newton’s third law requires that the electric forces on the two charges be the same, regardless of the equality or none quality of the charges. The solution of the example remains the same. The symmetry of the problem would be destroyed if the masses of the spheres were not the same. In this case, the strings would make different angle with the vertical and the problem would be more complicated. Page-5 q2 in accelerated motion.emerges only when we go beyond electrostatics and deal with time-dependent electromagnetic phenomena. We define the electric field due to the source charge at the location of the test charge to be the electric force on the test charge per unit charge. as in figure (a) the charge on the metallic sphere. When another charged object. If the test charge is great enough (q0 >> q0). which produces the electric field. Suppose we consider the force between two distant charges q1.The electric field The concept of a field was developed by Michael Faraday. however. Fe When using E q . qo (the test charge) enters this space. we must assume that the test charge q0 is small enough that it does not disturb the 0 charge distribution responsible for the electric field. the measurable quantity is the force on a charge which can be directly determined using Coulomb’s law . as in figure (b) the charge on the metallic sphere is redistributed and the ratio of the force to the test charge is different : (F’e / q’0 Fe/q0). An electric field is said to exist in the region of space around a charged object. That is. because of this redistribution of charge on the metallic sphere. If vanishingly small test charge q0 is placed near a uniformly charged metallic sphere. The direction of E. which is separate from the field produced by the test charge itself. the interaction between two charges is a two step process. Fe due to this field. The test charge is used to detect the electric field. Also. the electric field it sets up is different from the field it sets up in the presence of the much smaller test charge q0. For two charges. + q0 – – – – – – – (a) – – – – – – – – – + q0 >> q0 (b) Why concept of electric field necessary? Modern understanding of electric interaction between two charges is visualized in terms of the electric field concept. Q q0 + + + + + Fe E q 0 E + + + + + + ++ The vector E has the SI units of newtons per coulomb (N/C). the concept of electric field is convenient. The greatest Page-6 . note that the an electric field is a property of its source. Thus. The true physical significance of the concept of electric field. as shown figure. A charge produces an electric field around itself . but not really necessary. is the direction of the force a positive test charge experiences when placed in the field It is important to remember that Electric field E is produced by charge Q. Why then introduce this intermediate quantity called the electric field? When charges are stationary. Electric field in electrostatics is an elegant way of characterising the electrical environment of a system of charges. we say the test charge experiences an electric force. the presence of the test charge is not necessary for the field to exist. this field then exerts force on the other charge. remains uniformly distributed.. even though electric and magnetic fields can be detected only by their effects (forces) on charges. The notion of field elegantly accounts for the time delay. as in figure (c). which then propagate with the speed c. If some more charge are added. the force on the test charge is toward the source charge. the electric field generated by both together is simply the vector sum of the two separate fields. the force exerted by point charge q on the test charge q0 is : qq Fe k e 20 rˆ r q0 q + ^r F F P r q + (a) ^r P r (b) where rˆ is a unit vector directed from q toward q0. q0 E q + ^r P F (c) q ^r – P r (d) If q is negative. which each occupy three-dimensional space. They can also transport energy.speed with which a signal or information can go from one point to another is c. i. the speed of light. Field due to a point charge According to Coulomb’s law. as in figure (d) Superposition of electrostatic fields If we are dealing with many charges then electric field at a point p is the vector sum q E k e 2i rˆi i r1 where ri is the distance from the ith source charge q. there is no change to the terms that were already there. Page-7 . directed away from q. The two fields. If we know the electric fields generated by two different sets of charges separately. they are regarded as physical entities.e. It is precisely here that the notion of electric field (strictly. the electric field is said to satisfy the principle of superposition. are superimposed on one another. they evolve according to laws of their own.. By E = Fe/q0. Thus. the effect of any motion of q1 on q2 cannot arise instantaneously. not merely mathematical constructs. Thus concept of field is now among the central concepts in physics. to the point P and rˆi is a unit vector directed from qi toward P. However. so the electric field at P is directed toward the source charge. Because it has this property. the electric field created by q is : q E e k e 2 rˆ r The source charge sets up an electric field at point P. There will be some time delay between the effect (force on q2) and the cause (motion of q1). provided that the original charges do not move. Thus. The field picture is this: the accelerated motion of charge q1 produces electromagnetic waves. They have an independent dynamics of their own. more terms are added to the summation. reach q2 and cause a force on q2. electromagnetic field) is natural and very useful. The positive charge q1 = 24.00m-x – q3 F13 + q1 x | q1 | | q 3 | | q 2 | | q3 | = k e (3.00 C is at the origin.Illustrations Ex. x = –3m.00m x q2 Sol. as shown in figure. we find that the positive root is x = 1m.00 x ) 2 x2 Noting that ke and |q3| are common to both sides and so can be dropped. ke + F23 3. we solve for x and find that (3. This is another location at which the magnitudes of the forces on q3 are equal.00m. x x=+a (i) E on point charge is undefined The variation of along the x-axis. 3. and the resultant force acting on q3 is zero. What is the x coordinate of q3? y=± Ex. Ey = 2kqy ( y a 2 )3 / 2 E 2 Emax = –a /2 8 kq 27 a 2 a /2 y occurs at a 2 Three point charges lie along the x axis as shown in figure. Two identical positive point charges q are placed on the x-axis at x = – a and x = + a. Electric field directed along the positive x-axis is taken as positive (ii) The direction of electric field along the positive y-axis is taken as positive. but both forces are in the same direction at this location. (ii) Plot the variation of E along the y-axis.00 – x)2 |q2| = x2 |q1| This can be reduced to the following quadratic equation : solving this quadratic equation for x. E +q +q x=–a Sol. Page-8 .0 C is at x = 3. (i) Plot the variation of E along the x-axis. There is also a second root. the positive charge q2 = 6. will it return to equilibrium or be pulled further from equilibrium? That is. Problem solving tactics for calculating the electric field from continuous charge distributions Identify the type of charge distribution and compute the charge density or . Thus. The amount of charge dq. In this case. 3. This results in a net force to the right. the equilibrium is stable. 2. Note that if the charge is constrained to allowed to move up and down along y -axis in figure. Find electric field at point p which is at a distance z lying on the line perpendicular to the plane of the square passing through the centre of square. When released.} To learn stable. the equilibrium is unstable. if the charge is pulled upward (or downward) and released. q E k e 2 i rˆi r1 i Considering the charge distribution as continuous.. Four identical charges are fixed at the corners of a square of side a. F13 becomes larger and F23 becomes smaller. unstable and nuetral equilibrium. Divide the charge distribution into infinitesimal charges dq. will move back toward the equilibrium position and undergo oscillation Explain using graph of E vs x.Asking question Suppose charge q3 is constrained to move only along the x axis. each of which will act as a tiny point charge. the total field at P in the limit qi 0 is q lim E k e qi 0 2 i rˆi = k dq rˆ e 2 r i r1 OPTIONAL 1. i. p q q q Ans. it is pulled a very small distance along the x axis.e. Ex. is the equilibrium stable or unstable? Ans. If the charge is moved to the right. within a small element dl. q qz . From its initial position at x = 1m. in the same direction as the displacement. 2 2 q 0a 3 Electric field of a continuous charge distribution : The total electric field at P due to all elements in the charge distribution is approximately. dA or dV is dq = dl (charge distributed in length) dq = dA (charge distributed over a surface) dq = dV (charge distributed throughout a volume) Page-9 . dE is directed away for positive charge dq while directed towards dq for negative dq. y = x tan x dy = x sec2 d K dy cos dEx = (x sec ) 2 dEy = so Ex = P y K dy sin ( x sec ) 2 dy K (sin 2 – sin 1) x K (cos 1 – cos 2) x 1 and 2 are to be used with sign. 5. What is the nature of the electric field at such a point ? Ans. and E kel/a2. 6. If P is far from the rod (a >> ).2 = 3 4 /3 /4 P Page-10 . then in the denominator of the final expression for E can be neglected. 7. The electric field is obtained by summing over all the infinitesimal contributions. P dq=dx dx x a l k el a (l a ) Asking question Suppose we move to a point P very far away from the rod.4. Ey = For eg in this figure 1 = . Ex. The magnitude of dE is 1 dq dE = 4 2 0 r Vector dE is along radial line joining dq to P. Resolve the dE vector into its components. Identify any special symmetry features to show whether any component(s) of the field that are not cancelled by other components. Draw at point P the dE vector produced by the charge dq. Electric field due to finite rod at perpendicular distance x from the wire. Perform the indicated integration over limit of integration that include all the source charges. Write the distance r and any trigonometric factors in terms of given coordinates and parameters. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end y x E Ans. E dE = dq 4 r 2 0 8. A thin rod of length has a uniform positive charge per unit length . x For infinite wire E= (ii) O K x Ey = 0 Ex = 2 Electric field due to arc = charge distribution / unit length. /2 dE = 0 °r K ( R d) cos R2 dr R d 2K sin 2 E net = R Rd d Electric field due to the ring (at a point P lying a distance x from its center along the central axis perpendicular to the plane of the ring) dq + + + + + + + a r + + + + x P dEx + + + + dE dE The magnitude of the electric field at P due to the segment of charge dq is dq r2 This field has an x component dEx = dE cos along the x-axis and a component dE perpendicular to the x-axis. Find field at the centre due to arc.(i) Results: For semi infinite wire E x K = x E y = O K x 2 K at 45° with OP. So only the horizontal components are cancelled and only vertical components are added. The resultant field at P must lie along the x-axis because the perpendicular components of the field created by any charge element is canceled by the perpendicular component created by an element on the opposite side of the ring. dE = ke Page-11 . O R R (a) [Ans. what type of motion does it exhibit ? In the expression for the field due to a ring of charge. from Fe = qE . the motion will be simple harmonic. Assuming a curvature radius R to be considerably less than the length of the thread. Thus. we can integrate to obtain the total field at P k x ( x 2 ea 2 )3 / 2 dq Ex = = kex Q ( x a 2 )3 / 2 2 1 dq + + + + + a + r + x + + + + P dEx + + + + + + + + Ex. we let x << a. (b) E = 0 ] 40 R Suppose a negative charge is placed at the center of the ring in and displaced slightly by a distance x << a along x-axis. dE2 (b) 2 .1 + + + + + + + + + dE2 + + + + + dE1 + 2 kex dq x dq dEx = dE cos = k e 2 = 2 ( x a 2 )3 / 2 r r All elements of the ring make the same contribution to the field at P because they are all equidistant from this point. dE1 + 2 O Ex. When released. the force on a charge –q placed near the center of the ring is Ex = k e qQ x a3 Because this force has the form SHM. (a) E = Ans. find the magnitude of the electric field strength at the point O. Fx = – Always look for the condition of SHM as F –x Page-12 . a and b. + + + + + + + dE dE + + + A thread carrying a uniform charge per unit length has the configurations shown in Fig. which results in k eQ x a3 Thus. dq R r x P dr The ring of radius r and width dr shown in has a surface area equal to 2rdr. Find the interaction force between the ring and the thread. but for x = 0 the answer is not valid as thereis discontinuty at x = 0. E x O Page-13 . x is a constant R Ex = kex 2r dr (x 2 r 2 )3 / 2 0 x = 2k e 1 2 2 1/ 2 ( x R ) This result is valid for all values of x > 0 and x<0. The charge of the thread (per unit length) is equal to . The total charge of the ring is equal to q. Ex. Using this result of field due to the ring kex dEx = 2 2 3 / 2 (2r dr) (x r ) To obtain the total field at P. Ex = 2ke = 2 0 * Explain that the field created by a uniformly charged infinite sheet is same. thus. q [Ans. Let disk has radius R has a uniform surface charge density . A system consists of a thin charged wire ring of radius R and a very long uniformly charged thread oriented along the axis of the ring. the expression in bracket reduces to unity to give us the near-field approximation. F = 4 R ] 0 Electric field due to disk (at a point P that lies along the central perpendicular axis of the disk and a distance x from the center of the disk.Asking question What will be the changes in the result if the ring is non uniformly charged. The charge dq on this ring is 2r dr. we integrate this expression over the limits r = 0 to r = R. with one of its ends coinciding with the centre of the ring.. We can calculate the field close to the disk along the axis by assuming that R >> x . Relative density of field lines is inversely propotional to square of distance We can imagine two equal and small elements of area placed at points R and S normal to the field lines there. shows such a picture. Draw vectors pointing along the direction of the electric field with their lengths proportional to the strength of the field at each point. because it was contained in the length of the arrow? No. the field gets weaker and the density of field lines is less. Now the magnitude of the field is represented by the density of field lines. To understand the dependence of the field lines on the area. all pointing outwards from the point charge. each arrow indicates the electric field. In this figure. Since the magnitude of electric field at a point decreases inversely as the square of the distance of that point from the charge. i. Let the point charge be placed at the origin. Having drawn a certain set of field lines. Since the electric field decreases as the square of the distance from a point charge and the area enclosing the charge increases as the square of the distance. Away from the charge. Figure. Relative density of field lines represent magnitude of electric field Have we lost the information about the strength or magnitude of the field now.ELECTRIC FIELD LINES and FLUX Concept of field lines We have already studied electric field in the few lectures.e. the force acting on a unit positive charge. E is strong near the charge. We draw the figure on the plane of paper. the vector gets shorter as one goes away from the origin..e. resulting in wellseparated lines. in two dimensions but we live in three-dimensions. or rather the solid angle subtended by an area element. We thus get many field lines. The field lines crowd where the field is strong and are spaced apart where it is weak. so the density of field lines is more near the charge and the lines are closer. Connect the arrows pointing in one direction and the resulting figure represents a field line. It is the relative density of lines in different regions which is important. Figure shows a set of field lines. always pointing radially outward. i. But the number of lines is not important. So if one wishes to estimate the density of field lines. Another person may draw more lines. In fact. the relative density (i.e. Let us try to represent E due to a point charge pictorially. whatever may be the distance of the area from the charge.. an infinite number of lines can be drawn in any region. let us Page-14 . The number of field lines in our picture cutting the area elements is proportional to the magnitude of field at these points. placed at the tail of that arrow. closeness) of the field lines at different points indicates the relative strength of electric field at those points. one has to consider the number of lines per unit cross-sectional area. We started by saying that the field lines carry information about the direction of electric field at different points in space.. perpendicular to the lines. It is a vector quantity and can be represented as we represent vectors. the number of field lines crossing the enclosing area remains constant. The picture shows that the field at R is stronger than at S. in general. (ii) In a charge-free region. A field line is a space curve. at a distance r. for two points P1 and P2 at distances r1 and r2 from the charge. If there is a single charge. electric field lines can be taken to be continuous curves without any breaks. respectively. a generalization of angle to three dimensions. (iii) Two field lines can never cross each other.. q) give a vivid pictorial description of their mutual repulsion. Recall how a (plane) angle is defined in two-dimensions.. Let a small transverse line element l be placed at a distance r from a point O. in three-dimensions the solid angle subtended by a small perpendicular plane area S. while those around the configuration of two equal and opposite charges (q. which is absurd. a dipole.mathematical way of visualizing electric fields around charged configurations. cutting unit area element is therefore n / (r12 ) at P1 and n / (r22 ) at P2. the field lines are in 3-dimensional space. Since n and are common. a curve in three dimensions.An arrow on the curve is obviously necessary to specify the direction of electric field from the two possible directions indicated by a tangent to the curve. Then the angle subtended by l at O can be approximated as = l/r. they may start or end at infinity. show clearly the mutual attraction between the charges. In Fig. The more appropriate term is field lines (electric or magnetic) that we will use. Faraday called them lines of force. a curve drawn in such a way that the tangent to it at each point is in the direction of the net field at that point. An electric field line is. We know that in a given solid angle the number of radial field lines is the same. The field lines around a system of two positive charges (q. Properties of field lines: The field lines follow some important general properties: (i) Field lines start from positive charges and end at negative charges. This term is somewhat misleading. The number of lines (say n) cutting these area elements are the same.try to relate the area with the solid angle.) Page-15 . the strength of the field clearly has a 1 / r2 dependence.e. The number of field lines. –q). the field at the point of intersection will not have a unique direction. Electric field lines are thus a way of pictorially mapping the electric field around a configuration of charges. i. As mentioned earlier. can be written as = S/r2. Figure shows the field lines around some simple charge configurations. The field lines of a single positive charge are radially outward while those of a single negative charge are radially inward. (If they did. especially in case of magnetic fields. Likewise. respectively. the element of area subtending the solid angle is r12 at P1 and an element of area r22 at P2. Drawing field lines The picture of field lines was invented by Faraday to develop an intuitive non. though the figure shows them only in a plane. We should however note that there is no flow of a physically observable quantity unlike the case of liquid flow. Page-16 . through a small flat surface dS.(iv) Electrostatic field lines do not form any closed loops. the number of field lines crossing the area element will be smaller. No lines should pass through this point. i. but makes an angle with it.. For the case of the electric field. In the picture of electric field lines described above. i. Therefore. the number of field lines crossing S is proportional to ES cos .e. we saw that the number of field lines crossing a unit area. Illustration 1. we define an analogous quantity and call it electric flux. the lines should be radially outward. Therefore. the pattern should look like that of a single point charge of value (2Q – Q) = + Q.. When = 90°. This means that if we place a small planar element of area S normal to E at a point. the lines are radial and spherically symmetric. – 2Q and +Q. 4Q. Figure shows the sketch of field lines for two point charges 2Q and – Q. Figure (a) shows the incorrectly and (b)shows the correctly drawn field lines for a collection of four charges – Q. Now suppose we tilt the area element by angle . (c) Far field : Far from the system of charges. (d) Null point : There is one point at which E = 0. The rate of flow of liquid is given by the volume crossing the area per unit time v dS and represents the flux of liquid flowing across the plane. field lines will be parallel to S and will not cross it at all (Figure). the projected area in a plane perpendicular to v is v dS cos . the pattern must be symmetrical about the line joining the two charges. Therefore the flux going out of the surface dS is v . placed normal to the field at a point is a measure of the strength of electric field at that point. Clearly. E=0 2Q –Q • The pattern of field lines can be deduced by considering the following points: (a) Symmetry : For every point above the line joining the two charges there is an equivalent point below it.e. The projection of the area element normal to E is S cos. the number of field lines crossing it is proportional to ES. nˆ dS. -2Q -2Q -Q +4Q P (a) +4Q -Q -Q (b) -Q ELECTRIC FLUX Analogy with flow of water and concept of flux Consider flow of a liquid with velocity v. If the normal to the surface is not parallel to the direction of flow of liquid. its own field predominates. This follows from the conservative nature of electric field. (b) Near field : Very close to a charge. (e) Number of lines : Twice as many lines leave + 2Q as entre – Q. in a direction normal to the surface. to v. Thus. Page-17 . Thus the direction of a planar area vector is along its normal. For the case of a closed surface. The vector associated with every area element of a closed surface is taken to be in the direction of the outward normal. component of E along the normal to the area element times the magnitude of the area element. E times the projection of area normal to E. Notice we could look at the expression E S cos in two ways: E (S cos ) i. The direction of an area element is along its normal. this convention is very simple.. How to associate a vector to the area of a curved surface? We imagine dividing the surface into a large number of very small area elements. Notice one ambiguity here. But a normal can point in two directions. is proportional to the number of field lines cutting the area element. How to specify the direction of a planar area? Clearly. Vectorial definition of flux We now come to the definition of electric flux. Electric flux through an area element S is defined by = E. with the convention stated already. as explained before. If you hold it normal to the flow.. For example.Area as a vector The orientation of area element and not merely its magnitude is important in many contexts. maximum water will flow through it than if you hold it with some other orientation. Which direction do we choose as the direction of the vector associated with the area element? This problem is resolved by some convention appropriate to the given context.S = E S cos which. in a stream. The angle here is the angle between E and S. the normal to the plane specifies the orientation of the plane. Each small area element may be treated as planar and a vector associated with it. is the angle between E and the outward normal to the area element. the area element vector S at a point on a closed surface equals S nˆ where S is the magnitude of the area element and nˆ is a unit vector in the direction of outward normal at that point. This is the convention used in Fig. For a closed surface.e. or ES i. the amount of water flowing through a ring will naturally depend on how you hold the ring. The unit of electric flux is N C–1 m2. as seen before. This shows that an area element should be treated as a vector. It has a magnitude and also a direction.e. Thus. Gauss’s Law Let us again consider a positive point cahrge q located at the centre of a sphere of radius r. = E. Thus. Thus. That is. E is constant over the surface and given by E = keq / r2. in principle. to calculate the total flux through any given surface. E ·Ai = EAi from equation E = ke and from equation E = E · dA = E n dA we find that net flux through are gaussian surface is E = E · dA = E dA = E dA where we have moved E outside of the integral because. Hence. the dependence on r cancels. as shown in figure Gaussian surface r + q dA E q ˆ we know that the magnitude of the electric field everywhere on the surface r2 r of the sphere is E = keq/r2. Furthermore. we can write this equation in the form E = q E = 0 q Note from equation E = that the net flux through the spherical surface is proportional to the charge 0 inside. by symmetry. E is parallel to the vector Ai representing a local element of area Ai srrounding the surface point. Therefore. because the surface is spherical dA = A = 4r2.S = E S cos can be used. All we have to do is to divide the surface into small area elements. at each surface point. in the product of area and electric field. the total flux through a surface S is ~ E ·S The approximation sign is put because the electric field E is taken to be constant over the small area element. calculate the flux at each element and add them up. This is mathematically exact only when you take the limit S 0 and the sum in Eq. The field lines are directed radially outward and hence are perpendicular to the surface at every point on the surface.The basic definition of electric flux given by Eq. the net flux through the gaussian surface is k eq (4r2) = 4keq r2 we know that ke = 1/4 0. The flux is indepdent of the radius r because the area of the spherical surface is proportional to r2. whereas the electric field is proportional to 1/r2. Page-18 . ~ E ·S is written as an integral. Because the charge is not at the center. but surface S2 and S3 are not. we conclude that the net electric flux through a closed surface that surrounds no charge is zero. But still the flux through the spherical surface is q / 0 Now consider several closed surfaces surrounding a charge q. the flux that passes 0 through S1 has the value q / 0. flux is proportional to the number f electric field lines passing through a surface. as shown. As we discussed in the preceding section. q As you can see from this construction. Flux through closed surface for a charge kept outside the surface Now consider a point charge located outside a closed surface of arbitary shape. From equation E = . the situation does not possess enough symmetry to evaluate the electric field. Thefore. S3 S2 S1 q Surface S1 is spherical. The number of electric field lines entering the surface equals the number leaving the surface. dA E In this case. The construction shown in figure shows that the number of lines through S1 is equal to the number of lines through the nonspherical surfaces S2 and S3. Therefore. Page-19 .Asking question What if the charge in figure were not located at the center of the spherical gaussian surface ? Gaussian surface r + q Ans. any electric field line that enters the surface leaves the surface at another point. we conclude that the net flux through any closed surface surrounding a point charge q is given by q / 0 and is independent of the shape of that surface. the magnitude of E would vary over the surface of the sphere and the vector E would not be everywhere perpendicular to the surface. as shown in. the net flux through S is q1 / 0. S q•1 q•4 •q2 • q3 S S The surface S surrounds only one charge. we can express the flux through any closed surface as : E .. The flux through S due to charges q2. q1 . which states that the electric field due to many charges is the vector sum of the electric fields produced by the individual charges.. (iii) (iv) In situations when the surface is so chosen that there are some charges inside and some outside. which is a generalization of what we have just described. d A 1 2 where E is the total electric field at any point on the surface produced by the vector addition of the electric fields at that point due to the individual charges. the net flux through surface S” is zero because there is no charge inside this surface. just term ‘q’ in the law represents only total charge inside. all the electric field lines that enter S” at one point leave at another. We once again use the superposition principle. Here are the list of different types of the Gaussian surfaces to be chosen for a given charge distribution. Therefore. (ii) The q includes sum of all charges enclosed by the surface. Whenever we apply Gauss’s law we may choose a surface of any size and shape as our Gaussian surface.Gauss’ Law for multiple charges Let us calculate flux for many charges.). Page-20 . the E (whose flux appear in the equation) is due to all charges.. Gauss’s law. That is. Application of gauss law Definition of a Gaussian surface : While applying Gauss’s law we are interested in evaluating the integral E = E · dA The closed surface for which the flux is calculated is generally an imaginary or hypothetical surface. Finally. it can pass through a continuous charge distribution. d A ( E E .. states that the net flux through any closed surface is : q E . q3 and q4 outside it is zero because each electric field line that enters S at one point leaves it at another. But selecting a proper size and shape for a Gaussian surface is a key factor for determining flux. The gaussian surface should not pass through any discrete charge however. called a Gaussian surface. Some points to be emphasized about the gauss law : (i) It is true for any closed surface no matter what its shape is size. the net flux through it is (q2 + q3)/ 0. Notice that charge q4 does not contribute to the net flux through any of the surfaces because it is outside all of the surface. The surface S’ surrounds charges q2 and q3 . hence. Consider the system of charges shown in figure. d A in E = 0 where qin represents the net charge inside the surface and E represents the electric field at any point on the surface. hence. Flux through ABCD.Charge distribution Point charge Spherical charge distribution Line of charge Planar charge Ex Gaussian surface Spherical Spherical Cylindrical Cylindrical Electric field Radial Radial Radial Normal to surface Consider a cube of edge a. CDHG is zero because E is normal to area vector on these faces. Note that flux through faces ABFE. Let us enclose the charge q by a cubical gaussian surface with q at its centre. Find electric flux through square of side a. Flux through EFGH. kept in a uniform electric field of magnitude E. (E)ABCD = E(a2) cos 0° = + Ea2 Area vector and electric field vector are parallel to each other. q• a/2 • a By symmetry all face have equal flux Page-21 . BCGE. Net flux over all the six faces is (E) = (E)ABFE + (E)BCGF + (E)ADHE + (E)CDHG + (E)ABCD + (E)EFHG = 0 + 0 + 0 + 0 + ( – Ea2) + Ea2 =0 Ex. ADHE. dA3 3 D C A E B dA1 l G 1 dA2 x z E l 4 2 F dA Sol. due to charge placed at distance a/2 from centre of a square. a/2 Sol. (E)EFGH = E(a2) cos 180° = – Ea2 Area vector and electric field vector are opposite to each other. directed along x-axis as shown in Figure. The net flux is algebraic sum of the flux through all the faces of the cube. A point charge q is placed on the apex of a cone of semi-vertex angle . Therefore. 2 0 Consider a Gaussian sphere with its centre at the apex and radius the slant length of the cone. A q E = A · 0 0 Here. dA = (2r) R d r Rd x R C (b) = (2R sin ) R d as r = R sin = (2R2) sin d A = ( 2R 2 ) sin d 0 A = 2R2 (1 – acos ) A q The desired flux is E = A · 0 0 q (1 cos ) (2R 2 ) (1 cos ) q = · = 2 2 0 (4 R ) 0 Objective : Introduction to application of solid angle Page-22 . q = 6 0 Objective : Flux from a point charge is distributed symmetrically Ex. A0 = area of whole sphere = 4R2 and A = area of sphere below the base of the cone. Consider a differential ring of radius r and thickness dr. Show that the electric flux through the base of the cone is Sol. the flux through the base of the cone. q (1 cos ) . The flux through the whole sphere is q / 0. A charge is placed at the centre of a cylindrical surface.OPTIONAL Ex. Page-23 . this law only gives E along dA. 45° q Total flux linked can be divided into parts (i) Flux through lateral surface L (ii) Flux through end caps A If end cap substends solid angle at centre. (i) x > R Student will attemp it like this Possible field Q Let gaussian surface of r = x Q dA = E 0 Q E 4x2 = 0 Q E = 4 x 2 0 directions R x {Emphasize this point} Note : But this soln is incomplete because. Find total flux passing through lateral curved surface. and how do we know that E is along dA . Shell A charged shell with total charge Q distributed uniformly on the surface of shell find field at x. q L + 2A = 0 q A = × 4 0 Note : Solid angle is given by = 2 (1 – cos ) = q 0 1 2 1 2 4 2 1 q 1 2 = 0 2 q L = 2 0 ] Calculation of electric field using Gauss Law : I. 2R R Sol. Only direction possible is radial. so E can be taken out of integral (ii) Q E dA = 0 Q E 4x2 = 0 Q = E 40 x 2 ] x<R In this case gaussian surface does not include any charge [as charge in gaussian surface is zero] E · dA = 0 Student can interpret it in this manner E is 0 Q R x Gaussian surface Emphasize that E · dA = 0 may also means E to dA or E is 0 or E · dA = 0 Emphasize proper method E · dA = 0 By symmetry E is along dA Also E · dA 0 as all will be either +ve or –ve by symmetry E dA = 0 Now dA 0 E 0 Page-24 .Proper Logic Field can’t be along tangent no tangential component can exist because of symmetry of charge distribution. Q 0 Q 0 E · dA = E · dA = Also by “symmetry” field at each point an gaussian surface has same magnitude is same. Uniformly charged sphere with const volumetric charge density. OPTIONAL Ex. (i) field at outside point x > R By symmetry we see that tangentical components of E is cancelled and net field is radial only. Charge is distributed throughout a spherical region of space in such a manner that its volume charge density is given by = ar2. 4 R 3 E dA = 3 0 x R (iii) 4R 3 2 E 4x = 3 0 x > R . Page-25 . 4 R 3 E · dA = 3 0 Also there is uniform charge distribution E at every point has same magnitude and directed radially.II. 0 r R where a s is a constant. 4 R 3 E= 3 2 40 x E Gaussian surface Field at inside points x<R Gaussian surface 4 x 3 E · dA = 3 0 x Again by symmetry E is only along radial direction. Hence proved. Find the field at distance r from the center. 4 x 3 E dA = 3 0 4 x 3 E 4x2 = 3 3 0 x E = 3 0 x<R Hence proved. and to be uniform over the end caps of the cylinder.Y R dr r Z Sol. ar3 5 0 Ex. Area of sphere is 4r2. Page-26 . Infinite sheet. Field inside a cavity : Consider a spherically symmetric charge distribution of charge density . We can think of entire spherical charge to be made up of concentric shells. Therefore. Sol. Objective : (i) Cavity can be considered having equal amount of positive and negative charge (ii) Cavity of uniform charged non conducting sphere has uniform E. Field inside cavity can be obtained by super position of electric field due to complete sphere and electric field due to cavity part i. constant charge density We choose a small closed cylinder whose axis is perpendicular to plane as our gaussian surface. dv = 4r2dr X Since charge distribution is volumetric we choose a small volume element dV within the sphere. . Because the charge distribution is spherically symmetric. we choose a thin spherical shell of radius r and thickness dr. dV = 4r2dr R q = dV = (ar 2 ) (4r 2 dr ) 0 = 4 ar 5 5 1 4 E = 4 . From symmetry we expect E to be directed perpendicular to the plane on both sides as shown. r Contribution of complete sphere E sphere = 3 0 x Ecavity = – 3 0 r x Resultant field E = E sphere + E cavity l E = 3 ( r x ) 0 E inside cavity = 3 l 0 So field inside cavity is uniform. III.e. at the given point subtract the contribution of electric field due to cavity part from field of complete sphere. with a cavity inside it. E . by symmetry again E is only along dA . dA left cap + A . Hollow cylinder (infinite length) What type of charge distribution cylinder can have? (i) Field outside cylinder x > R Linear charge distribution and surface charge distribution Let us consider a coaxial cylindrical gaussian surface.Gaussion surface A E dA • E x A = E · d A 0 E . = E dA 0 right cap Now if the consider a plane through x. charge above and below is same. dA = 0 curved Thus. E = 2 0 Asking question : Why we should consider the cylindrical gaussian surface in front as well as behind the sheet? IV. dA + curved E . l = E · d A ε0 x By previous proved reason l l E x 2 x l = ε 0 E = 2x 0 (ii) Field inside cylinder x<R q enclosed = 0 E · dA = 0 Now E is uniform throughout and radial Also E is along dA in region x < R so this shows that E = 0 E 0 x<R 2 3 1 3 Page-27 . through lateral surface is zero as E is along x-axis and here dA is perpendicular to it. There is a repulsive force between the two charges and when they are separated by a distance r the magnitude of the force is given by equation qq1 F = 4 r 2 0 The force on q acts in the direction AB. The amount of work done when q moves from B to C. The potential energy of q depends only on its distance from q1 and not on the direction. It is natural to choose the potential energy to be zero at rf = . but this energy does not disappear. rf rf r qq1 1 f qq1 qq1 W = Fdr = r = 4 2 dr = 4 ri 0 0 ri 4 0 r ri 1 1 ri rf This work represents the difference in the electrical potential energy of the system of the two charges when q moves from B to C. the work done by the force is Fdr. Page-28 .e. Note : Electric force is a Central force. Central force : Any force satisfying | F | f (r ) i. but q may be moved. which may later be released : gravitational potential energy may. separated by a distance ri. If q moves a distance dr along the line BC. but if q and q1 have different signs the right-hand side is 0 i negative as the work must be done to pull the charges apart. Examples considered in class XI included the work done against the gravitational force in lifting a mass. It is stored as potential energy. The work done by a force was defined as Work done = force × distance moved by the point of application of force. Work done by electric charges The concepts of work and potential energy also apply when the forces are electrical. function of distance from a fixed point Fˆ rˆ i. directed along line joining All Central forces are conservative force. In both cases energy must be expended to do the work.e. and with this choice the total potential energy U of the two charges when they are at A and B. and against the restoring force of a spring when it is stretched. for example. is qq1 U = W = 4 r 0 i qq1 Equation U = W = 4 r is still valid. be released by allowing an object to fall. Consider the two positive charges q and q1 as shown in figure.ELECTRIC POTENTIAL The electrostatic Potential Energy The concepts of work and potential energy were discussed in class XI. A q1 ri q B C rf Lets say the charge q1 is fixed. U V = q = – 0 1. 4. This quantity U/q0 is called the electric potential (or simply the potential) V.. Just as with potential energy. then U is negative. the electric potential at any point in an electric field is : U V= q 0 Since potential energy is a scalar quantity so the electric potential also is a scalar quantity. Electric potential is a scalar characteristic of an electric field. we see that if q0 is positive. 5. (This fails for line charge & sheet). dl B A Units : Joule / coulomb = volt 2. Page-29 . E . Thus. It is a scalar quantity.The Electrostatic Potential The potential energy per unit charge U/q0 is independent of the value of q0 and has a value at every point in an electric field. independent of any charges that may be placed in the field. As work can be +ve or –ve.The most general reference point is infinite & potential at is assumed to be zero. We can calculate the change in the potential energy of the charge-field system from Equations : U = q0V = –q0Ed From this result. the charge-field system experiences a change in potential energy. We often take the value of the electric potential to be zero at some convenient point in an electric field. If the test charge is moved between two positions A and B in an electric field. E g Now suppose that a test charge q0 moves from A to B. Potential is defined as the change in potential energy of the system when a test charge is moved between the points divided by the test charge q0. Dimension : w M1L2T 2 = = M1L2T–3A–1 q AT 3. Electric fields lines always point in the direction of heighest decreasing electric potential A A d d q m B B can be explained using analogy with gravity. only differences in electric potential are meaningful. the electric potential can also be +ve or –ve. Q. The change in potential energy of the charge-field system for this process is (a) positive (b) negative (c) zero. 3. Calculate Potential difference between points A and B in front of Infinite charge sheet. this gives B V = – E ds Ed A The negative sign indicates that the electric potential at point B is lower than at point A . for which the field lines are radial and the equipotential are spherical. Direction of electric field at any point on equipotential surface is to the surface and in direction in which v is decreasing. This is easily proved by considering a small movement of a test charge on an equipotential surface. Spherical surfaces centred on q1 are equipotential surfaces. and it follows that the electric field does not have a component lying in the surface. W. Asking Questions Q. no matter what the distribution of charge.Ex. q1 Electric field is perpendicular to the equipotential surface. 1. that is. Page-30 . a negative charge is placed at A and then moved to B. to take a charge from one point on equipotential surface to other is zero. the field is perpendicular to the surface. Electric field is always perpendicular to equipotential surfaces. 4. Properties of equipotential surfaces. E B A Equipotential surfaces : For an isolated charge q1 the electrostatic potential at a distance r from q1 is q1 / (40r).1 In figure. This is obvious for a single charge. d s = – (E cos 0)ds = – E . If E 3ˆi 4ˆj N/C and potential at origin is zero find potential at (2. Ex. that is. VB < VA. The potential difference V = VB – VA is (a) positive (b) negative (c) zero. Two equipotential surface do not intersect each other. two points A and B are located within a region in which there is an electric field. d s B A A A Because E is constant. All points on the surface of a sphere of radius r are at the same potential. 2. B B VB – VA = V = – E . we can remove it from the integral sign .D.2). (if v = 0 at ).1) to (2. 4) and find work required to take a particle of charge –10C from (1. These are imaginary surfaces where potential of all points are equal.2 In figure. No work is done. 2 Find the direction of field at P.3 l • • Calculate shape of equipotential surfaces. where all potentials are calculated with same reference point. A B P C D 20V 10V 5V C is direction of E field Q. dq Potential on the axis of ring + + + + + + + a + + + + x P + + + (1) r + Potential at P using relation between E & V q x Eeff = 4 2 2 3/ 2 = 0 (r x ) Page-31 . EP. E . ER in order of magnitude. Equi Potential surfaces 10V 20V • • P A 30V • Q B 40V R C D ER > EQ > EP Q.1 Arrange ER. Potential due to combination of charges : (superposition principle) If a point is located in field of more than one charge then potential of that point is summation of individual potential with sign. Methods of calculation of potential (1) Find E in space and calculate potential at a point using V = . d s (2) Calculate potential due to elements and add them up.Asking question Q. * * Page-32 .d s = x (2) q 4 0 x x (r 2 x 2 )3 / 2 q dx = 4 0 r 2 x2 Potential at P using superposition q 1 dV = 40 dq 2 r x 0 2 q = 4 0 r 2 x2 Asking question Q.1 What happens if the ring is non uniformly charged? Q. V = 0 Q 2 R & Emax = 2 3 3 40 R 2 (iii) E is maximum at x = ± V E –R /2 R /2 x r Potential difference due to Infinite charged wire +ve r1 r2 Potential difference in moving from r1 r2. V = E eff . V= q (maximum) 40 r (ii) Infinite E = 0 .2 Draw graph of V versus r and relate it with the graph of E (i) Centre at x = 0 E=0 . r2 So V2 – V1 = 20r (–dr) r1 V2 – V1 = 2 (nr1 – nr2) 0 So here it is clear that V1 0 as r1 Also here as V 0 kdq kdq V = dV = will not work as when we write potential due to an element as we x x assume potential at infinite to be zero. 1 V r V r=Rr A fixed solid sphere of radius R is made of a material dielectric constant K = 1.Potential due to solid uniformly charged non conducting sphere R r 1 Q V = 4 0 r r>R 4 r3 3 V= + 40 r V= 0 R 4 x 2 dx 40 x r r<R 1 2 1 2 r2 q(3R 2 r 2 ) r R = 2 2 80 R 3 3 Results : (i) Centre r = 0 E=0 . From a distance x (x > R) from centre. V = 4 R 0 (Graph of field and potential due to shell) E E Inverse square ~ I/r2 Linear ~r O r a Inverse square ~ I/r2 Linear ~r O r a parabola 1 V r (iii) V r=Rr Ex. V= (ii) Surface r = R q E = 4 R 2 0 3 1 Q 2 40 R q . Uniform charge density . A bullet of mass m and having charge q is fired towards its centre. calculate min. velocity so it can cross the sphere. Finding E from V E V = – grad (V) Page-33 . Assume no other force acting on the bullet except the electrostatic force. . = – ( ˆi ˆj kˆ ) The electric potential varies in space according to the relation V = 3x + 4y. 3.E. Electric potential energy of a system Definition : It is the work done to assemble the charges from infinite separation to present configuration w/o change in K. ˆ ˆ ˆ i j k (x.. PE2 = P21 + PE23 useful for symmetric arrangements.. V V =1 . 1.I. A particle of mass 10Kg starts from rest from point (2.. 1) V V V = yz .. y) are in S. y. PEsys = Wi (ii) Find PE of each charge dues to field other charges. PEsys = PE1 PE 2 PE 3 .. 1.. x E = – ( ˆi Ex... Methods of calculation (i) Keep all charges at separation from each other and then bring them one by one in present configuration and calculate the work done.. A B q1 q2 q1 C Page-34 . Find the velocity of the particle when it crosses the x-axis.. units. 2mm/s] III..2) under the influence of this field.. The charge on the particle is +1C. = xz .. Concept of Self energy / potential energy of system energy required to create a system is called self energy.. =1 y z ˆj kˆ ) V = xyz find E at (1.. of any particle. V = x + y + z find E V =1 ... Assume V and (x. 2 Where PE1 = PE12 + PE13 + . [Ans. = xy y x z E = – ( yzˆi xzˆj xykˆ ) Q. z co-ordinate system) grad = y z x Ex. Shell Method I : let say that x charge has been brought and if further dx charge from inifinity. Find energy to the break. 1 U= 2 7 i 1 7 q Vi / r r 1 7 3k (q) 3kq k (q ) Vi / r = a a 2 a 3 r 1 as because of (–q) is also same. –q q q –q –q q –q a q kq 3ka 3ka 1 Work done to distroy = – U = . Q U= U= kx dx R 0 R Q KQ 2 2R Method II : Here V due to other charges is Q dq Q 1 kQ U = R dq 2 0 multiplied by U= 1 because every interaction counted two times 2 kQ 2 2R Page-35 .Method I Work done to bring A w1 = 0 Work done to bring B w2 = kq1q 2 r Work done to bring C w3 = kq1q 3 kq 2 q 3 + r r Self energy = w1 + w2 + w3 1 q (V VA / C + qB (VB / A VB / C ) + qC (VC/A + VC/B)] 2 A A/B Method II : U = 1 U= 2 n n q i Vi / r i 1 r 1 Ex.8q a 2 a 3 2 a Calculation of self energy due to continuous charge distributions 1. the system Sol. 3kQ12 3kQ 22 kQ1Q 2 U= 5R 1 5R 2 r Energy density of electric field * Energy / unit volume where field is 1 0E 2 2 This is the energy required to create electric field per unit volume. Find self energy of system r Q1R1 Sol. Q2R2 Because of three parts to create both sphere and then their interacles. 2 1 1 Q 2 U = 0 2 4x dx 2 4 x 0 R 1 k 2Q2 2 = 0 2 4x dx 2 x R 1 k 2Q 2 4 = 0 2 2 R kQ 2 2R Note : We don’t write self energy of point charge.2. = Page-36 . Ex. energy required to make field at x. Solid sphere Only by Method I : Let us consider a shell of radius x has ben created and further dx them created. Ans. Self energy of shell Now let us consider at x. R 4 / 3x 3 U = 4 x (4 x2 dx)) 0 0 3kQ 2 U= 5R Ex. 140. Electric potential due to dipole P r A –q O for d << r i. What amount of work has to be performed to slowly transfer the charge q from the point O through the orifice and into infinity? Ans. d• • –q • q d (i) Dipole moment : P = q d . BPe = r – d cos 2 AP = r + qB short dipole d cos 2 Page-37 . (v) All the distances are measured from centre of dipole.Ex. d is always taken reason -ve to +ve (ii) Axis of dipole : is line joining -q to +q (iii) Mid point of axis of dipole is centre of dipole (iv) bisector of line joining +q to -q is equatorial line of dipole.e. ) The inside and outside radii of the layer are equal to a and b respectively. Electric Dipole Definition : When two charges of equal magnitude and opposite sign are separated by a very small distance. centre O of a spherical uncharged conducting layer provided with a small orifice (Fig. then the arrangement is called electric dipole. Solid sphere :Here since E in sphere varies in different ways for x > R and x < R 2 R 1 x 4x 2dx + U = 0 2 3 0 0 U= 2 1 1 Q 2 2 0 40 x 2 4x dx R x 3kQ 2 5R {Optional} Irodov 3. A point charge q is located at the Fig. Page-38 . on axis for given distance r E E E • P =0 V +ve + = V –ve Formula notapplicable 1 2P E = 4 3 0 r 1 P V = 4 2 0 r E is in direction of P (ii) Equatorial line : Minimum potential and field at = 90° i.e. •P – + V=0 Find equivalent dipole moment of a ring having linear charge densities + and – on its two halves. on equitorial plane for given distance r. On the axis : Maximum potential and electric field at = 0 i.e. . E 1 P E = 4 3 0 r Ex.kq kq k q d cos ~ – = kq (d 2cos ) d d r2 d 2 2 r cos r cos r cos 2 2 4 k p . ˆr kp = 2 = 2 cos r r VP = Electric field due to dipole : E ( V ) E ˆ kp cos = – rˆ r r 2 r =– r • P – kP = 3 ( 2 cos rˆ sin ˆ ) r P 1 3 cos 2 (i) + where (r >> d) r3 tan where is angle of Enet with r 2 direction tan = 2P cos Er = 4 r 3 0 Er • 2kP cos kP sin ˆ rˆ + 3 r r3 1 E = 4 0 Enet P sin E = 4 r 3 0 . d 2 dt 2 •+ –• • – 2 PE d 2 =– I dt = E stable for small PE I + • t • unstable +• •– • Page-39 . is zero dU = – d 90° = – PE cos d –• 0 = – PE sin = – PE cos = – P·E (D) It will undergo SHM if turned by small angle PE sin = – I . It is given by U = –p. 2 E +• Potential energy in this posn. Note : • • • the potential energy is minimum at = 0 the potential energy is maximum at = the dipole is in stable equilibrium at = . or U = –p E cos U O /2 where is the angle between the dipole and the electric field vector. as shown in figure. P Electric dipole in uniform field : (A) (B) (C) Force F1 F2 = 0 Fnet 0 Torque +q E qE • E • • –q –qE l | | qE sin 2 2 In the vector form p E Potential Energy The work done by an external agent in rotating the dipole without change in its kinetic energy is stored an potential energy in the field and dipole system. There potential energy U as a function of the angle is plotted in figure.E. calculate electric intensity at the midpoint of the aperture. Page-40 . 2 dF = dq E = ( ds) 2 = 2 ds 0 0 So force per unit area on a charged conductor due to its own charge 2 dF 1 = = 0 E 2 2 0 ds 2 [as for a conductor E = ] 0 This force is called ‘mechanical force’ or electrostatic pressure.... ER + ED = and ER – ED = 0 0 Solving these for ER and ED : ER = ED = 2 0 i.. then for a charged spherical shell (or conductor) Eout = and Ein = 0 0 Now as for outside the shell both ED and ER will be directed outwards while inside ER will be outwards while ED inwards so that : Eout = ER + ED and Ein = ER – ED .. Note : As intensity on the disc (element) the to remainder is (/20).. (1) and (2). electric force on it will be.. field at the aperture will be (/20) directed outwards. ]. If ED and ER are the intensities due to disc and the remainder respectively at P.e. (2) And hence equating Eqs. ER ER P ED ED a [Sol. Consider the shell to be made up of a disc of radius b and the remainder..Electrostatic Pressure {Optional} If a small piece of radius b is removed from a charged spherical shell of radius a (>>b). assuming the density of charge to be . In the static situation. Conductors contain charge carriers. and move randomly in different directions. Lets keep a positive charge q charge near a neutral or acharged conductor then the electrons goes close to q. then the e¯ move such that they will create a stronger E int which will tend to cancel Eext. The positive ions made up of the nuclei and the bound electrons remain held in their fixed positions. the free charge carriers would experience force and drift. There may also be an external electrostatic field. the electric field is zero everywhere inside the conductor. these charge carriers are electrons. Further Explanation What happens if conductor is placed in an exterrnal field. These electrons are free within the metal but not free to leave the metal. when there is no current inside or on the surface of the conductor. Inside a conductor. As long as electric field is not zero. When the conductor is charged. In a metal. 1. The free electrons form a kind of ‘gas’. they collide with each other and with the ions. A conductor has free electrons. the excess charge can reside only on the surface in the static situation. neutral or charged. E net E ext E int 0 2. The interior of a conductor can have no excess charge in the static situation A neutral conductor has equal amounts of positive and negative charges in every small volume or surface element. electrostatic field is zero Consider a conductor. In the static situation. of free charges which move in random direction so the lattice becomes positively charged. + + E net E ext E int ¯ + ¯ ¯ ¯ ¯ Eint + Eext + ¯ + ¯ neutral conductor So an e¯ experience E net If Eint Eext . the free charges have so distributed themselves that the electric field is zero everywhere inside.CONDUCTORS A conductor means infinite no. Page-41 . the outer (valence) electrons part away from their atoms and are free to move. Electrostatic field is zero inside a conductor. This fact can be taken as the defining property of a conductor. E net has to be O instantaneously. The redistribution of electrons inside conductor takes place which generates an internal electric field E int .. This constitutes current and therefore energy conservation is not valid. it would have some non-zero component along the surface. therefore.Explanation This follows from the Gauss’s law. qenclosed = 0 Since the surface S can be made as small as you like. the free electrons experience a force in the opposite direction of the field and migrate to one side of the conductor as shown in figure (a). Consider any arbitrary volume element v inside a conductor. and any excess charge must reside at the surface. 3. If we consider any small gaussian surface inside Q E · dA = 0 [as E = 0] On the closed surface S bounding the volume element v. Page-42 . electrostatic field must be normal to the surface at every point If E were not normal to the surface. Note : Thus Solid “ conducting” sphere is same as a shell. electrostatic field is zero. E should have no tangential component. At the surface of a charged conductor. This means there is no net charge at any point inside the conductor. Thus electrostatic field at the surface of a charged conductor must be normal to the surface at every point. Free charges on the surface of the conductor would then experience force and move. + + Emphasize following facts from figure (i) Field is normal to surface conductor (ii) Field inside conductor is zero Further discussion When we place an ideal conductor in an electric field E. the volume v can be made vanishingly small. In the static situation.. + + + Sol. (For a conductor without any surface charge density. Draw the field near the infinite sheet. by Gauss’s law.) Asking question Consider a neutral conducting sphere placed near a infinite non conducting uniform sheet of charge.e. field is zero even at the surface. i. Hence. Thus the total electric flux through S is zero. there is no net charge enclosed by S. Note : You may emphasise again but qin = 0 does not imply that E = 0 from gauss law. the contribution to the total flux through the pill box comes only from the outside (circular) cross-section of the pill box. This equals ± ES (positive for > 0. the electrostatic field is zero. The pill box is partly inside and partly outside the surface of the conductor. an infinitesimal distance below the surface. Note: Also if some external field is present then charge distribution on a Solid conducting sphere and a conducting shell will be non uniform. negative for < 0). Just inside the surface. the electric field is zero at every point on this Gaussian surface because it is inside the conductor [figure (b)]. E may be considered constant and E and S are parallel or antiparallel. the field is normal to the surface with magnitude E. the electric field inside an ideal conductor is zero. The charge enclosed by the pill box is S. OPTIONAL : Electric field at the surface of a charged conductor E = nˆ 0 where is the surface charge density and nˆ is a unit vector normal to the surface in the outward direction. Therefore. choose a pill box (a short cylinder) as the Gaussian surface about any point P on the surface. since over the small area S. This charged distribution creates an electric field in a direction opposite to the applied field. just outside. Gauss’s law then implies that the net charge contained within the Gaussian surface is zero. By Gauss’s law Page-43 . in electrostatic equilibrium. as shown in figure. No free charge can exist anywhere within the electrostatic conductor. excess charge on an ideal isolated conductor must reside on the conductor’s surface. The redistribution of charge takes place till net field inside the conductor is zero. It has a small area of cross section S and negligible height. To derive the result. If we place a Gaussian surface. Thus.– – + – + + + + – E (a) (a) The accumulation of electrons leaves one side positively charged and the other negative. In electrostatic equilibrium. the result. Eq. electric field is normal to the surface inward. At portions where the surface is more curved. each conductor is characterised by a constant value of potential. Page-44 . Here the charge density also depends on external field and E out E ind . this means potential will be different for the surface and a point just outside the surface. If a charge is given to this conductor figure. the charge density will not be uniform on the entire surface. electric field is normal to the surface outward. Hence.ES = | | S 0 || E= 0 Including the fact that electric field is normal to the surface. That is. the charge density will be larger. A portion where the surface is more “flat” may be considered as part of a sphere of larger radius. the charge density will be larger where the radius of curvature is small. shape and charge configuration. For > 0. for < 0. E = nˆ . there is no potential difference between any two points inside or on the surface of the conductor. electric field normal to the surface exists. we get the vector relation. Charge density on a conductor’s surface in abscence of an external field + ++r + +++ ++ + + + + + + + + 1 r + R + + + + Now consider a single conductor with a nonspherical shape. More precisely. int = 0 ¯ • Q ¯ ¯ ¯ ¯ ¯ + • + + + + Q1 Note : So here charge Q1 must be distributed non uniformly unrelated to Radius of curvature. 0 which is true for both signs of . The charge density at such a portion will be smaller from equation (1). Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface Since E = 0 inside the conductor and has no tangential component on the surface. no work is done in moving a small test charge within the conductor and on its surface. In a system of conductors of arbitrary size. but this constant may differ from one conductor to the other. If the conductor is charged. 4. Thus the distribution of charge Q on surface of conductor is “Non-uniform 1 and R curvature Charged conductor in external field : Figure shows a charge Q placed infront of a neutral conductor. Potential due to induced charges is zero as centre C is equidistant from all induced charges Volume of solid conductor is equipotential volume Vc = Vp = Kq KQ l R Also at P VP = Vdue to q + Vdue to induced charges + Vdue to Q Kq KQ Kq KQ + = + + Vdue to induced charges l R x R Kq Kq – l x E external + E conductor + Einduced = 0 Resultant electric field inside material is zero Vinduced charge = Ex. is a general result. Find electric field due to induced charges at P Kq | E p | by induced charges = 2 x Cavity inside a conductor : Cavity is aplace surrounded from all sides by the conductor such that without touching the body we can’t reach cavity. But the vanishing of electric field in the (charge-free) cavity of a conductor is. A point charge q is kept at a distance l from neutral sphere of radius R. A remarkable result is that the electric field inside the cavity is zero. The effect can be made use of in protecting sensitive instruments from outside electrical influence. Find electric potential V at P. solid conducting sphere. all charges reside only on the outer surface of a conductor with cavity. with no charges inside the cavity. A related result is that even if the conductor is charged or charges are induced on a neutralconductor by an external field. Electrostatic shielding Consider a conductor with a cavity. Q P x – q + + + r C – – – R – + l Sol. This is known as electrostatic shielding. The proof of the result for the shell makes use ofthe spherical symmetry of the shell. any cavity in a conductor remains shielded from outside electric influence: thefield inside the cavity is always zero. as mentioned above.Ex. Figure gives a summary of the important electrostatic propertiesof a conductor. We have proved a simple case of this result already: the electric field inside a charged spherical shell is zero. Page-45 . whatever be the size and shape of the cavity and whatever be the charge on the conductor and the external fields in which it might be placed. Whatever be the charge and field configuration outside. Furthermore. we conclude that a cavity surrounded by conducting walls is a field-free region as along no charges are inside the cavity. The only way that this can be true for all paths is if E is zero everywhere in the cavity.1. Let us consider a gauesion surface just near to cavity. If there is no charge present inside the cavity than field inside it is zero. U V q = – 0 B E · ds A B VB – VA = – E · ds A Because VB – VA = 0. Page-46 . B• •A Let us assume that no charges are inside the cavity. E · dA = 0 q in = 0 Sol. Thus. the electric field inside the cavity must be zero regardless of the charge distribution on the outside surface of the conductor. the field in the cavity is zero even if an electric field exists outside the conductor. From this we don’t have proved that no charge resides on cavity but have proved that net charge on cavity surface is O. If there is some charge present here and there in cavity such that there sum total is zero then the following disscussion will not be valid) Proof wrong sol. and therefore any two points A and B on the surface of the cavity must be at the same potential. In this case. (This does not mean that we are implying “no net charge”. we use the fact that every point on the conductor is at the same electric potential. Now suppose a conductor of arbitrary shape contains a cavity as shown in figure. the integral of E · ds must be zero for all paths between any two points A and B on the conductor. Now imagine that a field E exists in the cavity and evaluate the potential difference VB – VA defined by equation. To prove this point. The electric field due to charges on the outer surface of conductor is zero for all the points inside the conductor seperately Consider a charged conductor having charge +q1 and Q is kept inside the cavity. equal and opposite charge appears on outer surface. As E in material of conductor is zero. E net and E A . EB andEC are fields due to charge A. E · dA = 0 qenclosed = 0 Thus equal and opposite charge is induced on the inner surface of cavity. + + + + B + + + A¯ ¯ ¯ ¯ •q ¯ ¯ ¯ •p + + C Gaussian surface + Proof: Let us consider a gaussian surface just outside cavity inside material of conductor. the induced charge -Q on the surface of the cavity as B and the charge on the surface of the conductor Q + q1 as C. 3. Lets call charge Q inside cavity as A. E net = E A + E B E C = 0 Now the electric field due to charges on the outer surface of conductor is zero for all the points inside the conductor seperately and the E B E A is zero seperately. In electrostatic equilibrium charge distribution will be as shown in figure +q –q q charge on inner surface of cavity is – q since material of conductor is initially neutral..2. A charge q is placed inside cavity. C B A Now field inside the conductor is. Figure shows a conductor with a cavity inside it.B and C inside the conductor. Equal and opposite charge is induced on the inner surface of cavity. Page-47 . and. it will only affect the charge distribution at B the charge distribution at C will remain unaffected. as Ex. A positive point charge +q is in the cavity at the centre of the sphere. Find field at P. [Symmetrical distribution] C ¯ ¯¯ B ¯ q•A¯ ¯ ¯ Note : When if we displace the charge q inside the cavity. uncharged spherical conductor has inner radius a and outer radius b. Then field inside the cavity is just due to charge A only as new field lines are already to B and E due to B = 0. assuming that V = 0 at r = . and cavityis having a charge q inside the cavity. Ex. RQ r2 E A + EB = O Ep = A hollow. •P + x + r •q¯A + + ¯ ¯¯ • R B + C + + + Sol. the conductor is neutral. Er V kq r2 a b kq r2 Z kq + kq – kq a b r kq b kq r a b r + – –+ a –+ +q –+ – + – b + Page-48 .Special case : When spherical cavity in present inside spherical conductor and charge is at the centre. Make the graph E and potential V(r) everywhere. E · ds = 0 4Q P A • B – 3Q C 5Q D E F 2Q+x –x 4Q–x 3Q-x x –3Q+x as through two lids E = 0 as part of conductor and through lateral surface E · ds = 0 qenclosed = 0 Important : (i) Thus facing surfaces have equal and opposite charges. What will be the new potential difference between the same two surfaces if the shell given a charge –3Q ? Shell Sphere ++ ++++ + + +Q b ++ + ++ ++ [Sol. 1 q Vin = VC = VS = 4 R 0 and 1 q Vout = 4 r 0 So if a and b are the radii of sphere and spherical shell respectively. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. In case of a charged conducting sphere. potential at their surfaces will be.. (i) 0 Now when the shell is given a charge (–3Q) the potential at its surface and also inside will change by V0 Page-49 . (4Q x ) / Area (2Q x ) / Area – =0 2 0 2 0 Ex. Find charge on all surfaces : Applying Gauss’ law on opposite faces.Ex..... (ii) Using this also prove that outer faces of the two last plates have equal charges. 1 Q 1 Q Vsphere = 4 a and Vshell = 4 b 0 0 And so according to given problem Q 1 1 V = Vsphere – Vshell = 4 a b .. [as no field because of equal is cancelled] 4Q – x – 2Q – x = 0 2x = 2Q x = Q.. Sol. Let the potential difference between the surface of the solid sphere and that of the outer surface of hollow shell be V.... q = (40a)V Now.1 3Q = 4 b 0 So that now 1 Q V’sphere = 4 + V0 0 a and 1 V’shell = 4 0 Q b V0 And hence Q 1 1 V’sphere – V’shell = 4 = V [from Eq. A metal sphere A of radius a is charged to potential V. charge will reside on outer surface of B and so the potential of B will be . potential everywhere inside and on the surface of shell will change by same amount and hence potential difference between sphere and shell will remain unchanged.e. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire ? Sol.e. (I) When two conductors are connected : When two conductors are connected charge redistributes on the connected conductors till their potential becomes equal Ex. if any charge is given to external shell. a VA = VB = (V) [< V as a < b] b Page-50 .. B b A a 1 q 1 40a a VB = 4 b = 4 V= V b b 0 0 Now as sphere A is inside B so its potential. the potential difference between sphere and shell will not change.(1)] b 0 a i. 1 q V = 4 a 0 a i. If the charge on sphere of radius a is q. when sphere A is enclosed by spherical conductor B and the two are connected by a wire. This is because by presence of charge on outer shell. Find charge on all four platess assume length ang breadth of plates as very large. Earthing We assume that (i) (ii) Earth is infinite resource and sink of charge so potential of earth will not change. 17Q 6 13Q 6 Earthing : Earthing means connecting it with Earth. The potential of earth is assumed to be “zero”. Page-51 . We get. d d 2d D A P • Q Sol.Ex. 3QA – 3QB = 2Q so 6QA = 17Q QB = 5Q – QA = 17Q 30 Q 17Q = 6 6 = II. So after earthing the charge on conductors vary so that potential of conductor becomes zero. 3Q –5Q 2Q Writing field at P we get let charge on 2nd and 4th be Q1 QA + QB = 5Q A B C QA D +ve QB Now since both surfaces have same potential Work done is zero from B to D Writing fields in region BC and CD and calculating the work. Figure shows three thin concentric spherical shells with initial charges as shown in the figure shell A & C are connected by wire and such that it does not touch B and shell B is earthed. Let finally charges be qA..... Sol. qB and qC. Sol. qA + qC = 2Q . (i) kq C kq B kq A + + =0 2k 2k 3k 3Q 3R qB qC q + + A = 0 .Ex. Now inner shell is earthed. Consider concentric spherical shells of negaligible thickness.. r1 [This implies that a charge + Q2 r has been transferred to earth leaving negative charge on A. (ii) 2 2 3 2Q 2R R Also kq C kq 2 kq A kq A kq B kq C + + = + + R 2R 3R 3R 3R 3R aC + qB qA qA qB qC + = + + 2 3 3 3 3 B A –Q . Let the charge be q1 Q2 r2 kq1 kQ 2 + r1 r2 = 0 r1 q1 = – Q2 r 2 Q1 r1 [new charge after earthing] This means that the inner shell is at zero potential and that electric field lines leave the outer shell and go to infinity but other electric field lines leave the outer shell and end on the inner shell. (iii) Page-52 ..] 2 Ex. Find charge on the inner shell.. Initial charges on these shells are Q1 & Q2.