Capacitors

CAPACITANCE 1CONCEPTUAL QUESTIONS Q. 1. The two charged conductors are touched mutually and then separated. What will be the charge on them? Ans. The charge on them will be divided in the ratio of their capacitances. We know that q = CV. When the charged conductors touch, they acquire the same potential. Hence, q ∝ C. Q. 2. The plates of a charged capacitor are connected to a voltmeter. If the plates of the capacitor are separated further, what will be the effect on the reading of the voltmeter? Ans. q V C = and 0 A C d ε = As the capacitor plates are separated, C decreases. Since charge on the plates remains the same, value of V increases. Hence, the reading of the voltmeter will increase. Q. 3. Any conducting object connected to earth is said to be grounded. Explain. Ans. The earth is an electron source or sink and is arbitrarily said to be at zero potential. A conducting body connected to earth is also at zero potential or “ground potential”. Alternatively, the capacitance of earth is so large that removal of electrons from it or supply of electrons to it makes no difference either in the charge or potential of earth. Q. 4. How does a spark discharge occur between two charged objects? Ans. The air between the two charged objects is subjected to an electric field. If the potential gradient in the intervening air column becomes high enough, the air is ionised and conducting path is formed for free electrons which move across to discharge the surfaces. Stored electric potential energy is dissipated as heat, light and sound. Q. 5. If a solid dielectric is placed between the plates of a capacitor, its capacitance increases. Is there any other advantage of solid dielectric? Ans. There are other two advantages of a solid dielectric. First, it helps in keeping the plates close together without touching. Secondly, we can now charge the capacitor to a high potential (V = q/ C). Q.6. Can you place a parallel plate capacitor (consisting of two plates) of 1 farad in your almirah? Ans. No. Suppose the two plates of the capacitor are separated by as small a distance as 1 mm. 0 A C d ε = or 3 8 2 12 0 (1 10 ) (1) 1.1 10 m 8.854 10 Cd A − − × = = × = × ε × This area is equal to the area of a square having each side more than 10 km. Modern technology, however, has permitted the construction of 1F capacitors of very moderate size. Q. 7. Given a solid metal sphere and a hollow metal sphere. Which will hold more charge? Both spheres are of same radius. Ans. Both the spheres will hold the same charge. It is because charge remains on the outer surface of a charged conductor (whether solid or hollow) and the spheres have equal surface areas. Q. 8. Two capacitors of capacitances 1 µF and 0.01 µF are charged to the same potential. Which will give more intense electric shock if touched? Ans. q = CV. Since V is constant, q ∝ C. It means that capacitor having large capacitance will store more charge. Hence, when 1µF capacitor is touched, the discharging current will be high and you will get more intense electric shock than in case of 0.01µF capacitor. Q. 9. Two spheres of different capacitances are charged to different potentials. They are then joined by a wire. Will total energy increase, decrease or remain the same? Ans. The two spheres are at different potentials. Therefore, when they are connected by a wire, there will be redistribution of charge (i.e., flow of charge through wire) till the two spheres attain the same potential. Due to the flow of charge through the connecting wire, some energy will be lost as heat. Hence, the total energy after connecting the spheres will decrease. Q. 10. Can there be potential difference between two adjacent conductors which carry the same positive charge? Ans. Yes. We know that V = q/C. The capacitance depends upon the dimensions of the conductor. If the two conductors are of different shapes and sizes, they will be charged to different potentials when given the same charge. Q. 11. What are the differences between conductors and dielectrics? 2 CAPACITANCE Ans. (i) Conductors have a large number of free electrons while dielectrics have practically no free electrons. (ii) When a conductor is placed in an external electric field, there is no electric field inside the conductor. However, when a dielectric is placed in an electric field, its molecules are polarised. The effect of this polarisation is to weaken the applied electric field within the dielectric. (iii) The dielectric constant of conductors is infinity while that of dielectrics is finite. (iv) The dielectric strength of conductors is zero while that of dielectrics is finite. (v) There is no limit to the current that a conductor can carry, provided that it can be kept cool enough. However, there is a limit to the electric flux that a dielectric will carry without breaking down. Q. 12. Show that capacitance of a metal plate A can be increased by bringing another metal plate B near A. Ans. The charge holding property of a conductor is called its capacitance. Fig. 5.47 shows an insulated metal plate A. Let positive charge be given to it till its potential becomes maximum. No further charge can be given to the plate as it would leak out. Now bring another insulated metal plate B near the plate A as shown in Fig. 5.47. The plate A will induce negative charge on the inner face of B and equal positive charge on the outer face. The induced positive charge tends to increase the potential of plate A while induced negative charge tends to decrease the potential of plate A. Since the induced negative charge is nearer to plate A than the induced positive charge, the net effect is that the potential of plate A decreases. Therefore, more charge can be given to plate A to raise its potential to maximum value. Thus the capacitance of conductor A is increased by bringing another uncharged conductor B near it. Q. 13. In the above question, show that capacitance of metal plate A can be further increased by earthing the plate B. Ans. If the plate B is earthed [See Fig. 5.48], the induced positive charge being free will flow to earth. However, induced negative charge remains since it is bound to the positive charge on plate A. As a result, the potential of plate A is sufficiently reduced. We can now give a large amount of charge to plate A to raise its potential to maximum value. We arrive at a very important conclusion that capacitance of an insulated conductor is increased by bringing near it an uncharged earthed conductor. Q. 14. Two identical metal plates are given charges q 1 and q 2 (q 2 < q 1 ) respectively. If they are now brought close to form a parallel-plate capacitor with capacitance C, what will be the potential difference between the plates ? Ans. Suppose A is the area of each plate. When the plates are held at a distance d, the capacitance of the parallel plate capacitor formed is 0 A C d ε = If E 1 and E 2 are the electric fields due to two plates, then net electric field E between the two plates is 1 2 1 2 1 2 0 0 0 / / 1 ( ) 2 2 2 q A q A E E E q q A = − = − = − ε ε ε ∴ P.D. between plates, 1 2 1 2 0 0 1 ( ) ( ) 2 2 d V Ed q q d q q A A = = − × = − ε ε 1 2 2 q q C − = 0 A C d ε   =     ∵ Q. 15. A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m 2 . Will he get electric shock if he touches the metal sheet next morning ? Ans. Yes, the man gets shock. The discharging current in the atmosphere will charge the aluminium sheet. When the man touches the sheet, the charge will flow to earth via the body of the man. Therefore, the man will get a shock. Q. 16. Capacitors P, Q and R have each a capacitance of C. A battery can charge the capacitor P to a potential difference of V. If after charging P, the battery is disconnected from it and the charged capacitor P is connected in the following separate instances to Q and R (i) to Q in parallel and (ii) to R in series, then what will be potential differences between the plates of P in the two instances ? Fig. 5.47 + + + + + + + + + + + + + + – – – – – – – + + + + + + + A B Fig. 5.48 + + + + + + + + + + + + + + – – – – – – – A B CAPACITANCE 3 Ans. When the capacitor P is charged to a potential difference V, it acquires charge q = CV. (i) When P is connected in parallel to Q Total capacitance = C + C = 2C Total charge = q + 0 = q ∴ Potential difference V′ across each capacitor is 2 2 2 q CV V V C C ′ = = = i.e. P.D. across P = V/2 (ii) When P is connected to R in series When P is connected to R in series, the circuit is incomplete and no charge is transferred from capacitor P. Hence potential difference across the plates of P remains V. Q. 17. The safest way to protect yourself from lightning is to be inside a car. Comment. Ans. We know that electric field inside a conductor is zero. Since the body of the car is a metal, the electric field inside it is zero. The discharging current due to lightning passes to earth through the metallic body of the car. Q.18. Five identical capacitor plates each of area A are arranged such that the adjacent plates are at a distance d apart. The plates are connected to a battery of V volts as shown in Fig. 5.49. What is the magnitude and nature of charge on plates 1 and 4 ? Ans. The system constitutes 4 capacitors in parallel across V volts. Since area of each plate is A and the separation between adjacent plates is d, all the four capacitors have the same capacitance C. The plate 1 acts as a positive plate of capacitor C 1 (of capacitance C ) formed between plates 1 and 2. ∴ Charge on plate 0 1 A CV V d ε = + = The plate 4 acts as a negative plate of both the capacitors C 3 (between plates 3 and 4) and C 4 (between plates 4 and 5). ∴ Charge on plate 0 2 4 ( ) A C C V V d − ε = − + = – + V 1 2 3 4 5 Fig. 5.49 CAPACITANCE 1 SOLVED EXAMPLE Example 1: Assuming earth to be an isolated conducting sphere of radius 6400 km, what is the capacitance of earth? Solution: Capacitance of earth, C = 4π ε 0 r Here, 4πε 0 9 1 ; 9 10 = × r = 6400 km = 6.4 × 10 6 m ∴ 6 9 6.4 10 9 10 C × = × = 0.711 × 10 –3 F = 711 µ µµ µµF This shows that farad is a very large unit of capacitance. Note: Since capacitance of earth is quite large, we choose earth as a level of zero potential for practical purposes. Think about this ! Example 2. An isolated sphere has a capacitance of 50pF. (i) Calculate its radius. (ii) How much charge should be placed on it to raise its potential to 10 4 V? Solution: (i) Capacitance of sphere, C = 4 π ε 0 r ∴ Radius of sphere, 9 12 2 0 1 (9 10 ) (50 10 ) 45 10 m 4 r C − − = × = × × × = × = πε 45cm (ii) Charge to be placed, q = CV = (50 × 10 –12 ) × 10 4 = 5 × 10 –7 C = 0.5 µ µµ µµC Example 3. Twenty seven spherical drops, each of radius 3mm and carrying 10 –12 C of charge are combined to form a single drop. Find the capacitance and potential of the bigger drop. Solution. Let r and R be the radii of smaller and bigger drops respectively. Volume of bigger drop = 27 × Volume of smaller drop or 3 3 4 4 27 3 3 R r π = × π or R = 3r = 3 × 3 = 9 mm = 9 × 10 –3 m Capacitance of bigger drop, 3 12 0 9 1 4 9 10 10 F 9 10 C R − − = πε = × × = = × 1pF Since charge is conserved, the charge on the bigger drop is 27 × 10 –12 C. ∴ Potential of bigger drop, 12 12 27 10 10 q V C − − × = = = 27V Example 4: A spherical capacitor has an inner sphere of radius 9 cm and an outer sphere of radius 10cm. The outer sphere is earthed. Assume there is air in the space between the spheres. What is the capacitance of the capacitor? Solution: Capacitance of the spherical capacitor is 0 4 ( ) A B B A r r C r r πε = − Here r B = 10 cm = 0.1 m; r A = 9 cm = 0.09 m; 4π ε 0 9 1 9 10 = × ∴ 9 0.09 0.1 9 10 (0.1 0.09) C × = × − = 100 × 10 –12 F = 100 pF Example 5. The thickness of air layer between two coatings of a spherical capacitor is 2 cm. The capacitor has the same capacitance as the capacitance of sphere of 1.2m diameter. Find the radii of its surfaces. Solution: Given : 0 0 4 4 A B B A r r R r r πε = πε − A B B A r r R r r ∴ = − Here r B – r A = 2 cm and R = 1.2/2 = 0.6m = 60 cm ∴ 60 2 A B r r = or r A r B = 120 cm Now (r B + r A ) 2 = (r B – r A ) 2 + 4 r A r B = (2) 2 + 4 × 120 = 484 ∴ 484 22cm B A r r + = = Since r B – r A = 2cm and r B + r A = 22cm, r B = 12cm ; r A = 10 cm 2 CAPACITANCE Example 6. The plates of a parallel plate air capacitor are separated by a distance of 1 mm. What must be the plate area if the capacitance of the capacitor is to be 1F? Solution: The capacitance of a parallel plate air capacitor is given by; 0 A C d ε = Here d = 1 mm = 10 –3 m; A = ?; C = 1F ∴ 3 12 0 1 10 8.854 10 Cd A − − × = = ε × = 1.1 × 10 8 m 2 Note the enormous magnitude of plate area required to have a capacitance of 1F. This shows that farad is a very large unit of capacitance. Example 7: What distance apart should the two plates each of area 0.2 m × 0.1 m of a parallel plate air capacitor be placed in order to have the same capacitance as a spherical conductor of radius 0.5m? Solution: Area of plate, A = 0.2 × 0.1 = 0.02 m 2 Radius of sphere, r = 0.5 m For parallel plate capacitor, C = ε 0 A/d For spherical conductor, C = 4π ε 0 r Since the capacitance of the two capacitors is the same, ∴ 0 A d ε = 4πε 0 r or 0.02 4 4 0.5 A d r = = π π × = 3.18 × 10 –3 m = 3.18 mm Example 8. Calculate the capacitance of a parallel plate air capacitor of plate area 30 m2; the plates being separated by a dielectric 2 mm thick and of relative permittivity 6. If the electric field strength between the plates is 500 V/mm, calculate the charge on each plate. Solution: Capacitance, 12 0 3 (8.854 10 ) (6) (30) 2 10 K A C d − − ε × = = × = 0.797 × 10 –6 F = 0.797 µ µµ µµF P.D. across plates, V = E × d = 500 × 2 = 1000 volts ∴ Charge on each plate, q = CV = (0.797 × 10 –6 )1000 = 0.797 × 10 –3 C = 0.797 mC Example 9. A p.d. of 10 kV is applied to the terminals of a capacitor consisting of two parallel plates, each having an area of 0.01 m 2 separated by a dielectric 1 mm thick. The resulting capacitance of the arrangement is 300pF. Calculate (i) charge on each plate, (ii) electric flux density, (iii) potential gradient, and (iv) relative permittivity of the dielectric. Solution: C =300 pF = 300 × 10 –12 F; V = 10 kV = 10 × 10 3 volts (i) Charge on each plate, q = CV = (300 × 10 –12 ) (10 × 10 3 ) = 3 × 10 –6 C (ii) Electric flux density, 3 3 10 0.01 q A × σ = = = –4 2 3 × 10 C/m (iii) Potential gradient, 3 3 10 10 1 10 V E d − × = = = × 7 10 V/m (iv) Electric intensity, 0 E K σ = ε ∴ Relative permittivity, 4 12 7 0 3 10 8.854 10 10 K E − − σ × = = = ε × × 3.39 Example 10. A parallel plate capacitor is to be designed with a voltage rating 1kV using a material of dielectric constant 3 and dielectric strength of 107 Vm –1 . What minimum area of the plate is required to have a capacitance of 50 pF? Solution. For reasons of safety, the electric field E between the plates should not exceed 10% of the dielectric strength of the dielectric i.e. E = 10% of 10 7 = 10 6 Vm –1 . Now V E d = ∴ 3 3 6 1 10 10 m 10 V d d − × = = = ∴ Plate area 12 3 12 (50 10 ) 10 8.85 10 3 A − − − × × = = × × -3 2 1.9 × 10 m CAPACITANCE 3 Example 11. Two parallel plate air capacitors have their plate areas 100 cm 2 and 500 cm 2 respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is 0.5 mm, what is the distance between the plates of the second capacitor? Solution: Let us denote the first capacitor by suffix 1 and second capacitor by suffix 2. Since the two capacitors have the same charge and potential, their capacitances (C = q/V) are equal i.e. C 1 = C 2 . ∴ 0 1 0 2 1 2 A A d d ε ε = ∴ 2 2 1 1 A d d A = Here A 2 = 500 cm 2 ; A 1 = 100 cm 2 ; d 1 = 0.5 mm = 0.05 cm ∴ 2 500 0.05 100 d = × = 0.25cm Example 12. Three capacitors have capacitances of 0.5 µF, 0.3 µF and 0.2 µF respectively. They are first connected to have maximum capacitance and then connected to have minimum capacitance. Find the ratio of maximum capacitance to minimum capacitance. Solution: (i) For maximum capacitance, all the capacitors will have to be connected in parallel. C P = 0.5 + 0.3 + 0.2 = 1 µF (ii) For minimum capacitance, all the capacitors will have to be connected in series. 1 1 1 1 31 0.5 0.3 0.2 3 S C = + + = or C S = 3/31 µF ∴ P S C C = 31 3 Example 13. Two capacitors of capacitance 15 µF and 20 µF are connected in series to a 600 V d.c. supply. Find (i) charge on each capacitor, (ii) p.d. across each capacitor. Solution: (i) Equivalent capacitance, 1 2 1 2 15 20 8.57 F 15 20 S C C C C C × = = = µ + + In series connection, charge on each capacitor is the same. ∴ Charge on each capacitor, q = C S V = (8.57 × 10 –6 ) × 600 = 5.14 × 10 –3 C (ii) P.D. across 15 µF capacitor 3 6 1 5.14 10 15 10 q C − − × = = = × 342.7V P.D. across 20 µF capacitor 3 6 2 5.14 10 20 10 q C − − × = = = × 257V Example 14. The total capacitance of two capacitors is 4µF when connected in series and 18 µF when connected in parallel. Find the capacitance of each capacitor. Solution: Let C 1 and C 2 be the unknown capacitances. Then, C 1 + C 2 =18 ...(i) when in parallel 1 2 1 2 4 C C C C = + ...(ii) when in series Multiplying eqs. (i) and (ii), C 1 C 2 = 72 Now 2 1 2 1 2 1 2 ( ) 4 C C C C C C − = + + 2 (18) 4 72 6 = − × = ± ... (iii) Solving eqs. (i) and (iii), we get, C 1 = 12 µ µµ µµF or 6 µ µµ µµF; C 2 = 6 µ µµ µµF or 12 µ µµ µµF Example 15. In the circuit shown in Fig. 5.13, the total charge is 750 µC. Find the values of V 1 , V and C 2 . Solution: Voltage across 6 1 1 6 1 (750 10 ) , 15 10 q C V C − − × = = = × 50V Applied voltage, V = V 1 + V 2 = 50 + 20 = 70 V Charge on C 3 = C 3 V 2 = (8 × 10 –6 ) 20 = 160 × 10 –6 = 160 µC ∴ Charge on C 2 = 750 – 160 = 590 µC 4 CAPACITANCE Fig. 5. 13 V + _ V 1 C 2 C = 1 15 F µ V = 2 20 V C = 3 8 F µ ∴ Capacitance of 6 2 590 10 20 C − × = = 29.5 × 10 –6 F = 29.5 µ µµ µµF Example 16: Obtain the equivalent capacitance for the network shown in Fig. 5.14. For 300 V d.c. supply, determine the charge and voltage across each capacitor. Fig. 5.14 C 1 C 2 C 4 C 3 200 pF 300 V 200 pF 100 pF 100 pF + _ Fig. 5.15 300 V + _ C 3 C 2 C 4 C 1 200 pF 100 pF 100 pF 200 pF A B C Solution: Equivalent Capacitance. The above network can be redrawn as shown in Fig. 5.15. The equivalent capacitance C′ of series-connected capacitors C 2 and C 3 is 2 3 2 3 200 200 100pF 200 200 C C C C C × × ′ = = = + + The equivalent capacitance of parallel combination of C′ (= 100 pF) and C 1 is C BC = C′ + C 1 = 100 + 100 = 200 pF The entire circuit now reduces to two capacitors C 4 and C BC (= 200 pF) in series. ∴ Equivalent capacitance of the network is 4 4 100 200 100 200 BC BC C C C C C × × = = = + + 200 pF 3 Charges and p.d. on various Capacitors Total charge, q = CV = 12 8 200 10 300 2 10 C 3 − − | | × × = × | \ . ∴ Charge on C 4 = 2 × 10 –8 C ∴ P.D. across C 4 , V 4 = 8 12 4 2 10 100 10 q C − − × = = × 200V P.D. between B and C, V BC = 300 – 200 = 100 V Charge on C 1 , q 1 = C 1 V BC = (100 × 10 –12 ) × 100 = 10 –8 C P.D. across C 1 , V 1 = V BC = 100 V P.D. across C 2 = P.D. across C 3 = 100/2 = 50 V Charge on C 2 = Charge on C 3 = Total charge – Charge on C 1 = (2 × 10 –8 ) – (10 –8 ) = 10 –8 C CAPACITANCE 5 Example 17: Fig. 5.16 shows a network of four capacitors. Determine the equivalent capacitance between points A and B. If a 10V battery is connected between A and B, how much total charge will be stored on the capacitors. Fig. 5.16 C 3 = 9 F µ C 2 = 9 F µ C 1 = 3 F µ C 4 = 9 F µ A B Fig. 5. 17 C 1 C 2 C 4 C 3 A B Solution: The given network is equivalent to the network shown in Fig. 5.17. The equivalent capacitance C′ of the series connected capacitors C 2 , C 3 and C 4 is given by; 2 3 4 1 1 1 1 1 1 1 1 9 9 9 3 C C C C = + + = + + = ′ ∴ C′ = 3 µF Between points A and B, we now have capacitors C 1 and C′ (= 3 µF) in parallel. Therefore, the equivalent capacitance C AB between points A and B of the network is C AB = C 1 + C′ = 3 + 3 = 6 µ µµ µµF Total charge stored on the capacitors is q = C AB × V = (6 × 10 –6 ) × 10 = 60 × 10 –6 C = 60 µ µµ µµC Example 18: Calculate the equivalent capacitance between points A and B in Fig. 5.18. Solution: Refer to Fig. 5.18. It is clear that one plate of each capacitor is connected to point A (plate 1, plate 4, and plate 5). Similarly, other plate of each capacitor (plate 2, plate 3 and plate 6) is connected to point B. Therefore, the three capacitors are in parallel. Hence, equivalent capacitance between A and B is C AB = C 1 + C 2 + C 3 Example 19: In the circuit shown in Fig. 5.19, find (i) the equivalent capacitance between A and D and (ii) the charge on 12 µF capacitor. Solution: (i) C AB = 10 µF Fig. 5. 19 8 F µ 8 F µ 12 F µ 10 F µ 400 V 8 F µ 8 F µ 8 F µ A B C D + _ C BC = 8 8 8 8 | | × | + \ . + (8) + (8) = 4 + 8 + 8 = 20 µF 10 20 20 F 10 20 3 AB BC AC AB BC C C C C C × × = = = µ + + C CD = 8 + 12 = 20 µF ∴ (20/ 3) 20 (20/ 3) 20 AD C × = = µ + 5 F Fig. 5.18 C 1 C 2 C 3 1 3 5 2 4 6 A B 6 CAPACITANCE (ii) Total charge, q = C AD × V = (5 × 10 –6 ) × 400 = 0.002C The total charge will divide between the parallel capacitors connected in the branch CD. P.D. between C and D, 6 0.002 100V 20 10 CD CD q V C − = = = × ∴ Charge on 12 µF capacitor = (12 × 10 –6 ) × 100 = 1.2 × 10 –3 C = 1.2 µ µµ µµC Example 20: In the network shown in Fig. 5.20 (i), C 1 = C 2 = C 3 = C 4 = 8 µF and C 5 = 10 µF. Find the equivalent capacitance between points A and B. Fig. 5. 20 C 3 C 3 C 3 C 5 C 5 C 4 C 4 C 4 C 2 C 2 C 2 C 1 C 1 C 1 A D C B ( ) ii ( ) iii ( ) i A A B B C C D D Solution: A little reflection shows that circuit of Fig. 5.20 (i) can be drawn as shown in Fig. 5.20 (ii). We find that the circuit is a Wheatstone bridge. Since the product of opposite arms of the bridge are equal (C 1 C 4 = C 2 C 3 because C 1 = C 2 = C 3 = C 4 ), the bridge is balanced. It means that points C and D are at the same potential. Therefore, there will be no charge on capacitor C 5 . Hence, this capacitor is ineffective and can be removed from the circuit as shown in Fig. 5.20 (iii). Referring to Fig. 5.20 (iii), the equivalent capacitance C′ of the series connected capacitors C 1 and C 2 is 1 2 1 2 8 8 4 F 8 8 C C C C C × ′ = = = µ + + The equivalent capacitance C″ of series connected capacitors C 3 and C 4 [See Fig. 5.20 (iii)]. 3 4 1 2 8 8 4 F 8 8 C C C C C × ′′ = = = µ + + Now C AB = C′ || C″ = 4 || 4 = 4 + 4 = 8 µ µµ µµF Example 21: A mica dielectric parallel plate capacitor has 21 plates, each having an effective area of 5 cm 2 and each separated by a gap of 0.005mm. Find the capacitance of the capacitor. Take the relative permittivity of mica as 6. Solution: For a multiplate capacitor, capacitance is given by; 0 ( 1) K A C n d ε = − 12 4 3 (8.854 10 ) 6 (5 10 ) (21 1) 0.005 10 − − − × × × × = − × = 0.1062 × 10 –6 F = 0.1062 µ µµ µµF Example 22: A parallel plate capacitor has three similar parallel plates. Find the ratio of capacitance when the inner plate is mid-way between the outers to the capacitance when inner plate is three times as near one plate as the other. Solution: Fig. 5.21 (i). shows the condition when the inner plate is mid-way between the outer plates. The arrangement is equivalent to two capacitors in parallel. Capacitance of this capacitor, 0 0 0 1 4 / 2 / 2 K A K A K A C d d d ε ε ε = + = Fig. 5.21 ( ) ii ( ) i d/2 3 /4 d d/2 d/4 Fig. 5.21 (ii) shows the condition when the inner plate is three times as near as one plate as the other. CAPACITANCE 7 Capacitance C 2 of this capacitor, 0 0 0 2 16 / 4 3 / 4 3 K A K A K A C d d d ε ε ε = + = ∴ C 1 /C 2 = 0.75 Example 23: Given some capacitors of 0.1µF capable of withstanding 15V. Calculate the number of capacitors needed if it is desired to obtain a capacitance of 0.1µF for use in a circuit involving 60V. Solution: Capacitance of each capacitor, C = 0.1 µF Voltage rating of each capacitor, V C = 15 V Supply voltage, V = 60 V Since each capacitor can withstand 15 V, the number of capacitors to be connected in series = 60/15 = 4. Capacitance of 4 series-connected capacitors, C S = C/4 = 0.1/4 = 0.025 µF. Since it is desired to have a total capacitance of 0.1 µF, number of such rows in parallel = C/C S = 0.1/0.025 = 4. ∴ Total number of capacitors = 4 × 4 = 16 Fig. 5.22 shows the arrangement of capacitors. Example 24: A cylindrical capacitor has two co-axial cylinders of length 20 cm and radii 15.4 cm and 15 cm respectively. The relative permittivity of the insulation is 5. If a p.d. of 5000V is maintained between the two cylinders, determine (i) capacitance of cylindrical capacitor and (ii) potential of inner cylinder. Solution: (i) The capacitance of a cylindrical capacitor is 9 9 10 10 5 0.2 10 F = 10 41.4log ( / ) 41.4log (15.4/15) K l C b a − − × = × × = 2.11 × 10 –9 F (ii) Since the outer cylinder is earthed, the potential of the inner cylinder is equal to p.d. between the two cylinders, i.e., potential of inner cylinder = 5000V. Example 25: A 5µF capacitor is charged to a p.d. of 100V and then connected to an uncharged 3 µF capacitor. Calculate p.d. across the capacitors. Solution: Charge on 5 µF capacitor, q = CV = (5 × 10 –6 ) × 100 = 0.0005 C When the two capacitors are connected through a wire, the total capacitance C P = 5 + 3 = 8 µF. The charge 0.0005 C is distributed between the two capacitors to have a common p.d. ∴ P.D. across capacitors 6 0.0005 8 10 P q C − = = × = 62.5 V Example 26: Two capacitors of capacitance 4 µF and 6 µF respectively are connected in series across a p.d. of 250V. The capacitors are disconnected from the supply and are reconnected in parallel with each other. Calculate the new p.d. and charge on each capacitor. Solution: In series-connected capacitors, p.d.s across the capacitors are in the inverse ratio of their capacitances. ∴ P.D. across 4 µF capacitor 6 250 150V 4 6 = × = + Charge on 4 µF capacitor = (4 × 10 –6 ) × 150 = 0.0006 C Since the capacitors are connected in series, charge on each capacitor is the same. ∴ Charge on both capacitors = 2 × 0.0006 = 0.0012 C Parallel connection. When the capacitors are connected in parallel, the total capacitance C P = 4 + 6 = 10 µF. The total charge 0.0012 C is distributed between the capacitors to have a common p.d. ∴ P.D. across capacitors 6 Total charge 0.0012 10 10 P C − = = × = 120 V Charge on 4 µF capacitor = (4 × 10 –6 ) × 120 = 480 × 10 –6 = 480 µ µµ µµC Charge on 6 µF capacitor = (6 × 10 –6 ) × 120 = 720 × 10 –6 = 720 µ µµ µµC Example 27: Two capacitors A and B are connected in series across a 200V d.c. supply. The p.d. across A is 120V. This p.d. is increased to 140V when a 3µF capacitor is connected in parallel with B. Calculate the capacitance of A and B. Solution: Let C 1 and C 2 µF be the capacitances of the capacitors A and B respectively. When the capacitors are connected in series [See Fig. 5.31 (i)], charge on each capacitor is the same. Fig. 5.22 C C C C + _ 60 V 8 CAPACITANCE ∴ C 1 × 120 = C 2 × 80 or C 2 = 1.5 C 1 ...(i) When a 3 µF capacitor is connected in parallel with B [See Fig. 5.31 (ii)], the combined capacitance of this parallel branch is (C 2 + 3). Thus the circuit shown in Fig. 5.31 (ii) can be thought of as a series circuit consisting of capacitances C 1 and (C 2 + 3) connected in series. ∴ C 1 × 140 = (C 2 + 3) 60 or 7C 1 – 3C 2 = 9 ...(ii) Fig. 5. 31 140 V 60 V 120 V 80 V C 2 C 1 A B + + _ _ 200 V 200 V ( ) ii ( ) i A B C 1 C 2 3 F µ Solving eqs. (i) and (ii), we have, C 1 = 3.6 µ µµ µµF; C2 = 5.4 µ µµ µµF Example 28: A 16 µF capacitor is charged to 100V. After being disconnected, it is immediately connected to an uncharged capacitor of 4 µF. Determine (i) the p.d. across the combination (ii) the electrostatic energies before and after the capacitors are connected. Solution: C 1 = 16 µF; C 2 = 4 µF Before joining Charge on 16 µF capacitor, q = C 1 V 1 = (16 × 10 –6 ) × 100 = 1.6 × 10 –3 C Energy stored, 2 6 2 1 1 1 1 1 (16 10 ) 100 2 2 U C V − = = × × = 0.08 J After joining When the capacitors are connected through a wire, the total capacitance, C p = C 1 + C 2 = 16 + 4 = 20 µF. The charge 1.6 × 10 –3 C distributes between the two capacitors to have a common p.d. of V volts. ∴ P.D. across capacitors, 3 6 1.6 10 20 10 P q V C − − × = = × = 80 V Energy stored, 2 2 1 1 2 2 P U C V = = (20 × 10 –6 ) × (80) 2 = 0.064 J It may be noted that there is a loss of energy. This is primarily due to the heat dissipated in the conductor connecting the capacitors. Example 29: A metal sphere 4 m in diameter is charged to a potential of 3 MV. Calculate the heat generated when the sphere is earthed through a long resistance wire. Solution: Potential at the surface of sphere, 9 9 10 q V r = × ∴ Charge on sphere, 6 3 9 9 (3 10 ) 2 0.67 10 C 9 10 9 10 V r q − × × × = = = × × × Energy stored in sphere 3 6 1 1 (0.67 10 ) (3 10 ) 2 2 qV − = = × × × = 1005 J When the sphere is earthed, stored energy will be dissipated as heat in the resistance wire. Example 30. A parallel plate capacitor is connected to a 12 V battery. The charge on the capacitor is 1.35 × 10 –10 C. If the plate separation is decreased to half, find the extra charge given by the battery. Solution. When the plate separation is decreased to half, the capacitance (C = ε 0 A/d) becomes twice. Therefore, the charge on the capacitor (q′ = C′V)becomes twice. The extra charge is supplied by the battery. Extra charge supplied by battery = q′ – q = 2q – q = q = 1.35 × 10 –10 C Example 31. A parallel plate 100 µF capacitor is charged to 500V. If the distance between the plates is halved, what will be the new potential difference between the plates and what will be the change in the new stored energy? CAPACITANCE 9 Solution: C = 100µF = 100 × 10 –6 F = 10 –4 F; V = 500 volts When plate separation is decreased to half, the new capacitance C′ becomes twice i.e. C′ = 2C. Since the capacitor is not connected to the battery, the charge on the capacitor remains the same. The potential difference between the plates must decrease to maintain the same charge. ∴ q = CV = C′V′ or 500 2 2 2 CV CV V V C C ′ = = = = = ′ 250 volts New stored energy 2 2 1 1 (2 ) 2 2 2 V C V C | | ′ ′ = = | \ . 2 2 1 1 1 2 2 2 2 CV CV | | = = | \ . 4 2 1 1 10 (500) 2 2 − = × × = 6.25J Example 32. Fig. 5.32 shows a circuit for a camera flash. A 2000 µF capacitor is charged by 1.5V cell. When a flash is required, the energy stored in the capacitor is made to discharge through a discharge tube in 0.1 ms giving a powerful flash. Calculate the energy stored in the capacitor and power of the flash. Solution: Energy stored in the capacitor is 2 6 2 1 1 (2000 10 ) (1.5) 2 2 U CV − = = × × × = 2.25 × 10 –3 J In order to produce the flash, the capacitor discharges in 0.1 ms (= 0.1 × 10 –3 s). ∴ Power of flash 3 3 2.25 10 Time 0.1 10 U − − × = = = × 22.5W Example 33: A parallel plate capacitor is partially filled with an ebonite plate of thickness 6 mm. The area of the plates of the capacitor is 2 × 10 –3 m 2 and the distance between them is 0.01m. The dielectric constant for ebonite is 3. Calculate the capacitance of the capacitor. Solution: Capacitance of the capacitor, 0 1 (1 ) A C d t K ε = − − Here A = 2 × 10 –3 m 2 ; d = 0.01 m; t = 6 × 10 –3 m; K = 3 ∴ 12 3 3 (8.854 10 ) 2 10 0.01 6 10 (1 1/ 3) C − − − × × × = − × − = 2.95 × 10 –12 F Example 34: A parallel plate capacitor has plate area of 2 m 2 spaced by three layers of different dielectric materials. The relative permittivities are 2, 4, 6 and thicknesses are 0.5, 1.5 and 0.3 mm respectively. Calculate the capacitance of the capacitor. Solution: Capacitance of capacitor, 0 3 1 2 1 2 3 A C t t t K K K ε = + + 12 3 3 3 (8.854 10 ) 2 0.5 10 1.5 10 0.3 10 2 4 6 − − − − × × = × × × + + = 0.026 × 10 –6 F Example 35: A capacitor is composed of two plates separated by 3 mm dielectric of relative permittivity 4. An additional piece of insulation 5 mm thick is now inserted between the plates. If the capacitor has now capacitance one-third of its original capacitance, find the relative permittivity of the additional dielectric. Solution: Fig. 5.40 (i) and Fig. 5.40 (ii) respectively show the two cases. For the first case, 0 1 0 4 K A A C d d ε ε × × = = ...(i) For the second case, 0 0 3 3 1 2 1 2 2 3 3 10 5 10 4 A A C t t K K K − − ε ε = = × × + + ...(ii) – + 1.5 V 2000 F µ Discharge Tube Electronic Trigger Fig. 5.32 10 CAPACITANCE ( ) ii 3 mm 5 mm K 2 = ? K 1 = 4 ( ) i 3 mm K 1 = 4 Fig. 5.40 Dividing eq. (i) by eq. (ii), 2 4 3 5 3 3 4 K | | = + | \ . ∴ K 2 = 3.33 Example 36: An air capacitor has two parallel plates of 1500 cm 2 area and held 5 mm apart. If a dielectric slab of area 1500 cm 2 , thickness 2 mm and relative permittivity 3 is now introduced between the plates, what must be the new separation between the plates to bring the capacitance to the original value. Solution: When a dielectric slab of thickness t (t < d) is introduced between the plates of air capacitor, the capacitance is given by; 0 ( / ) A C d t t K ε = − − ...(i) If the medium were totally air, the capacitance would have been 0 0 A C d ε = ...(ii) An inspection of eqs. (i) and (ii) reveals that with the introduction of the dielectric slab between the plates of air capacitor, its capacitance increases. The distance between the plates is effectively reduced by t – (t/K). In order to bring the capacitance to the original value, the plates must be separated by this much distance in air. ∴ New separation between plates = d + ( t – t /K) = 5 + ( 2 –2/3) = 6.33 mm Example 37: The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4 mm. It is charged to 200V and the charging battery is removed. Now a dielectric slab (K = 4) of thickness 2 mm is introduced between the plates. Find (i) final charge on each plate, (ii) p.d. between the plates, (iii) final energy in the capacitor, and (iv) energy loss. Solution: The capacitance of parallel plate air capacitor is given by; 0 0 A C d ε = ...(i) When a dielectric slab of thickness t is placed between the plates of the capacitor, capacitance is given by; 0 m A C t d t K ε = − + ...(ii) ∴ 0 ( / ) 4 2 (2/ 4) 5 4 8 m C d t t K C d − + − + = = = Before Introduction of the Dielectric Slab When air capacitor is charged to 200 V and then battery is removed, charge (q) on each plate is q = C 0 × V 0 = (50 × 10 –12 ) × 200 = 10 –8 C After Introduction of the Dielectric Slab (i) Since the battery is removed before the introduction of the dielectric slab, the charge on capacitor plates will remain the same after the introduction of the dielectric slab. Final charge on each plate = q = 10 –8 C (ii) When dielectric slab is placed between the plates of the capacitor, its capacitance increases to C m and p.d. between plates decreases to V. q = C 0 V 0 = C m V ∴ 0 0 5 200 8 m C V V C | | = = × = | \ . 125V CAPACITANCE 11 Note that p.d. between the plates decreases. This is in agreement with theory. (iii) Final energy stored in the capacitor, 8 1 1 (10 ) 125 2 2 U qV − = = × = –7 6.25 × 10 J (iv) Initial stored energy, U 0 = 1 2 qV 0 ; Final stored energy, U = 1 2 qV ∴ Energy loss = U 0 – U = 1 2 q (V 0 – V) = 1 2 (10 –8 ) (200 – 125) = 3.75 × 10 –7 J This loss of energy will be apparent to the person who introduced the slab. The capacitor would exert a tiny force on the slab and would do work on it equal to 3.75 × 10 –7 J. Note: Initial stored energy, U 0 = 1 2 qV 0 = 1 2 (10 –8 ) × 200 = 10 – 6 J Final stored energy U = 6.25 × 10 –7 J Had the slab been 4 mm thick (= distance between plates), the energy stored in the capacitor would have been smaller by 1/K i.e. U 0 /K. Thus we arrive at a very important conclusion that final energy after the slab filling entire space is introduced is smaller by 1/K. Example 38. A parallel plate capacitor having plate separation of 3mm possesses a capacitance of 17.7 pF. The capacitor is connected to a 100V supply. Explain what would happen if a 3 mm thick mica sheet of dielectric constant 6 were inserted between the plates (i) while the voltage supply remains connected (ii) after the supply was disconnected. Solution. Capacitance of capacitor, C = 17.7 pF ; Dielectric constant of mica, K = 6 (i) When voltage supply remains connected When mica sheet is inserted between the plates of the capacitor, the capacitance becomes K times i.e. C′ = KC = 6 × 17.7 = 106.2 pF The p.d. between the plates of the capacitor remains equal to 100 V. Since C = q/V and V is same but C has increased, the charge on capacitor must increase i.e. Charge on capacitor, q′ = C V = 106.2 × 10 –12 × 100 = 1.06 × 10 –8 C The extra charge is supplied by the battery. (ii) After the voltage supply is disconnected C′ = KC = 6 × 17.7 = 106.2 pF Charge on capacitor, q = CV = 17.7 × 10 –12 × 100 = 1.77 × 10 –9 C Since the supply is disconnected, the charge on the plates remains the same. Because the capacitance (C = q/V) has increased, the potential difference across the plates must decrease to maintain the same charge. 9 12 1.77 10 106.2 10 q V C − − × ′ = = = ′ × 16.67V Example 39. In a Van de Graaff generator, the shell electrode is at 25 × 105 V. The dielectric strength of the gas surrounding the electrode is 5 × 10 7 V/m. Calculate the minimum radius of the spherical shell. Solution. Electric potential V of a charged shell is 0 1 4 q V r = πε Electric field at the surface of a charged shell is 2 0 1 4 q E r = πε ∴ V r E = or 5 2 7 25 10 5 10 m 5 10 V r E − × = = = × = × 5cm 12 CAPACITANCE MORE NUMERICAL PROBLEMS 1. A capacitor of 20 µF and charged to 500 V is connected in parallel with another capacitor of 10 µF charged to 200 V. Find the common potential. [Roorkee] [400 V] Hint. Charge on one capacitor, q 1 = C 1 V 1 = (20 × 10 –6 ) × 500 = 0.01 C Charge on second capacitor, q 2 = C 2 V 2 = (10 × 10 –6 ) × 200 = 0.002 C Total charge on capacitors, q = q 1 + q 2 = 0.01 + 0.002 = 0.012 C Total capacitance, C = C 1 + C 2 = (20 × 10 –6 ) + (10 × 10 –6 ) = 30 × 10 –6 F ∴ Common potential, 6 0.012 400V 30 10 q V C − = = = × 2. Five equal capacitors connected in series have a resultant capacitance of 4 µF. What is the total energy stored in these when connected in parallel and charged to 400 V ? [E.A.M.C.E.T. 91] [8 J] Hint. Suppose C µF is the capacitance of each capacitor. Since 5 (= n) capacitors are connected in series, 4 C n = or C = 4 n = 4 × 5 = 20 µF When the capacitors are connected in parallel, then equivalent capacitance C′ is C′ = 5 × 20 = 100 µF = 100 × 10 –6 F Energy stored is given by; 1 2 U = C′ V 2 = 1 2 × (100 × 10 –6 ) × (400) 2 = 8 J 3. Find the length of the paper used in a capacitor of capacitance 2 µF if the dielectric constant of the paper is 2.5 and its width and thickness are 50 mm and 0.05 mm respectively. [90 m] Hint. 0 K A C d ε = Here C = 2 µF = 2 × 10 –6 F; K = 2.5; d = 0.05 mm = 0.05 × 10 –3 m ∴ 6 3 2 12 0 (2 10 ) (0.05 10 ) 4.5m 8.85 10 2.5 Cd A K − − − × × × = = = ε × × ∴ 3 Area 4.5 Length = m 90m Width 50 10 − = = × 4. A 5 µF capacitor is fully charged across a 12 V battery and connected to an uncharged capacitor. The voltage across it is found to be 3 V. What is the capacity of the uncharged capacitor ? [E.A.M.C.E.T.] [15 µ µµ µµF] Hint. The common potential V after connection is 1 1 2 2 1 2 C V C V V C C + = + Here C 1 = 5 µF; V = 3 volts; V 1 = 12 volts, V 2 = 0; C 2 = ? ∴ 2 2 5 12 0 3 5 C C × + × = + ∴ C 2 = 15µF 5. A parallel-plate capacitor has plates of dimensions 2 cm × 3 cm. The plates are separated by a 1 mm thickness of paper. (i) Find the capacitance of the paper capacitor. The dielectric constant of paper is 3.7. (ii) What is the maximum charge that can be placed on the capacitor ? The dielectric strength of paper is 16 × 16 6 V/m. [(i) 19.6 × 10 –12 F (ii) 0.31 µ µµ µµC] Hint. (i) 0 K A C d ε = Here ε 0 = 8.85 × 10 –12 C 2 N –1 m –2 ; K = 3.7; A = 6 10 –4 m 2 ; d = 1 10 –3 m 12 4 12 3 (8.85 10 ) (3.7) (6 10 ) 19.6 10 F 1 10 C − − − − × × × × = = × × (ii) Since the thickness of the paper is 1 mm, the maximum voltage that can be applied before breakdown occurs is CAPACITANCE 13 V max = E max × d Here E max = 16 × 10 6 V/m; d = 1 mm = 1 × 10 –3 m ∴ Maximum charge that can be placed on capacitor is q max = C V max = (19.6 × 10 –12 ) × (16 × 10 3 ) = 0.31 × 10 –6 C = 0.31 µC 6. A capacitor of capacitance C 1 = 1 µF withstands the maximum voltage V 1 = 6 kV while another capacitance C 2 = 2 µF withstands the maximum voltage V 2 = 4 kV. What maximum voltage will the system of these two capacitors withstand if they are connected in series ? [M.N.R. 1992] [9 kV] Hint. The maximum charge q 1 and q 2 that can be placed on C 1 and C 2 are q 1 = C 1 V 1 = (1 × 10 –6 ) × (6 × 10 3 ) = 6 × 10 –3 C q 2 = C 2 V 2 = (2 × 10 –6 ) × (4 × 10 3 ) = 8 × 10 –3 C The charge on capacitor C 1 should not exceed 6 × 10 –3 C. Therefore, when capacitors are connected in series, the maximum charge that can be placed on the capacitors is 6 × 10 –3 C ( = q 1 ). ∴ 3 3 1 1 6 6 1 2 6 10 6 10 1 10 2 10 max q q V C C − − − − × × = + = + × × = 6 × 103 + 3 × 10 3 = 10 3 (6 + 3) = 9 × 10 3 V = 9 kV 7. A parallel-plate capacitor is charged with a battery to a charge q 0 as shown in Fig. 5.70 (i). The battery is then removed and the space between the plates is filled with a dielectric of dielectric constant K. Find the energy stored in the capacitor before and after the dielectric is inserted. 2 2 0 0 0 0 ; 2 2 q q C K C Hint. Energy stored in the capacitor in the absence of dielectric is 2 0 0 0 1 2 U C V = Since V 0 = q 0 /C 0 , this can be expressed as : 2 0 0 0 2 q U C = ... (i) Eq. (i) gives the energy stored in the capacitor in the absence of dielectric. After the battery is removed and the dielectric is inserted between the plates, charge on the capacitor remains the same. But the capacitance of the capacitor is increased K times i.e., new capacitance is C′ = K C 0 [See Fig 5.70 (ii)]. Fig. 5. 70 q 0 q 0 KC 0 V 0 + + _ _ ( ) i ( ) ii C 0 ∴ Energy stored in the capacitor after insertion of dielectric is 2 2 0 0 0 0 2 2 q q U U C K C K = = = ′ or 0 U U K = ... (ii) Since K > 1, we find that final energy is less than the initial energy by the factor 1/K. How will you account for “missing energy” ? When the dielectric is inserted into the capacitor, it gets pulled into the device. The external agent must do negative work to keep the dielectric from accelerating. This work is simply = U 0 – U. Alternately, the positive work done by the system = U 0 – U. 8. Suppose in the above problem, the capacitor is kept connected with the battery and then dielectric is inserted between the plates. What will be the change in charge, the capacitance, the potential difference, the electric field and the stored energy ? Hint. Since the battery remains connected, the potential difference V 0 will remain unchanged. As a result, electric field ( = V 0 /d) will also remain unchanged. 14 CAPACITANCE The capacitance C 0 will increase to C = K C 0 The charge will also increase to q = K q 0 as explained below. q 0 = C 0 V 0 ; q = C V 0 = KC 0 V 0 = K q 0 Initial stored energy, 2 0 0 0 1 2 U C V = Final stored energy, 2 2 0 0 0 0 0 0 1 1 2 2 U C V K C V KU = = = ∴ U = KU 0 Note that stored energy is increased K times. Will any work be done in inserting the dielectric? The answer is yes. In this case, the work will be done by the battery. The battery not only gives the increased energy to the capacitor but also provides the necessary energy for inserting the dielectric. 9. A parallel plate capacitor is maintained at a certain potential difference. When a 3mm slab is introduced between the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab. [M.N.R.] [K = 5] Hint. The capacitance of parallel-plate capacitor in air is 0 A C d ε = ... (i) With the introduction of slab of thickness t, the new capacitance is 0 (1 1/ ) A C d t K ε ′ = ′ − − ... (ii) Now the charge (q = CV) remains the same in the two cases. ∴ 0 0 (1 1/ ) A A d d t K ε ε = ′ − − or d = d′ – t (1 – 1/K ) Here d ′ = d + 2.4 × 10 –3 m ; t = 3 mm = 3 × 10 –3 m ∴ d = d + 2.4 × 10 –3 – 3 × 10 –3 1 1 K | | − | \ . or 2.4 × 10 –3 = 3 × 10 –3 1 1 K | | − | \ . ∴ K = 5 10. The capacitance of a variable radio capacitor can be changed from 50 pF to 950 pF by turning the dial from 0° to 180°. With dial set at 180°, the capacitor is connected to 400 V battery. After charging, the capacitor is disconnected from the battery and the dial is turned at 0°. (i) What is the potential difference across the capacitor when the dial reads 0° ? (ii) How much work is required to turn the dial if friction is neglected ? [M.N.R.] [(i) 7600 V (ii) 1.37 × 10 –3 J] Hint. (i) With dial at 0°, the capacitance of the capacitor is C 1 = 50 pF = 50 × 10 –12 F With dial at 180°, the capacitance of the capacitor is C 2 = 950 pF = 950 × 10 –12 F P.D. across C 2 , V 2 = 400 V ∴ Charge on C 2 , q = C 2 V 2 = (950 × 10 –12 ) × 400 = 380 × 10 –9 C When the battery is disconnected, charge q remains the same. Suppose V 1 is the potential difference across the capacitor when the dial reads 0°. ∴ q = C 1 V 1 or 9 1 12 1 380 10 7600V 50 10 q V C − − × = = = × (ii) Work required = Gain in energy of the capacitor 2 2 1 1 2 2 1 1 2 2 C V C V = − 12 2 12 2 1 1 50 10 (7600) 950 10 (400) 2 2 − − = × × × − × × × = 1.37 × 10 –3 J CAPACITANCE 15 11. A 90 pF capacitor is connected to a 12 V battery and is charged to 12 V. How many electrons are transferred from one plate to the other ? [6.9 × 109] Hint. q = CV = (90 × 10 –12 ) × (12) = 1.1 × 10 –9 C Now q = ne ∴ Number of electrons transferred is 9 9 19 1.1 10 6.9 10 1.6 10 q n e − − × = = = × × 12. If C 1 = 20 µF, C 2 = 30 µF and C 3 = 15 µF and the insulated plate of C 1 be at a potential of 90 V, one plate of C 3 potential of 90 V, one plate of C 3 being earthed, what is the potential difference between the plates of C 2 , three capacitors being connected in series ? [20 V] Hint. Fig. 5.71 shows the conditions of the problem. The equivalent capacitance of this series combination is given by; 1 2 3 1 1 1 1 1 1 1 3 20 30 15 20 C C C C = + + = + + = 6 20 20 F 10 F 3 3 C − = µ = × P.D. across series combination, V = 90 – 0 = 90 V Total charge, q = CV = 20 3 × 10 –6 ×90 = 6 × 10 –4 C ∴ P.D. across 4 2 2 6 2 6 10 , 20V 30 10 q C V C − − × = = = × 13. A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (i) Determine the capacitance of the capacitor. (ii) What is the potential of the inner sphere ? (iii) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Why is the capacitance smaller in the latter case ? [(i) 5.55 × 10 –9 F (ii) 4.5 × 102 V (iii) 1.33 × 10 –11 F] Hint. (i) Capacitance of spherical capacitor is given by; 0 4 ab C K b a = πε − Here 0 1 4πε = 9 × 10 9 C 2 N –1 m –2 ; K = 32; b = 0.13 m; a = 0.12 m ∴ 9 9 32 0.12 0.13 5.55 10 F 0.13 0.12 9 10 C − × = × = × − × (ii) Potential of inner sphere is 6 2 9 2.5 10 4.5 10 V 5.55 10 q V C − − × = = = × × (iii) Capacitance of isolated sphere is 11 0 9 0.12 4 1.33 10 F 9 10 r − = πε = = × × Note that capacitance of an isolated sphere is much smaller than that of the concentric spheres. It is because in case of concentric spheres, the total potential is distributed over two spheres and the potential difference between the two spheres becomes smaller. Since capacitance (C = q/V) is inversely proportional to potential difference, a spherical capacitor has large capacitance. 14. N drops of mercury of equal radii and possessing equal charges combine to form a big drop. What is the charge, capacitance and potential of the bigger drop ? [(i) Q = Nq (ii) C = c N 1/3 (iii) V = v N 2/3 ] Hint. Let q, v and c be the charge, potential and capacitance of the individual small drop. The corresponding quantities for the bigger drop are Q, V and C. Charge on bigger drop = N × charge on small drop ∴ Q = Nq The capacitance of a spherical drop is proportional to the radius. 16 CAPACITANCE ∴ C R c r = Since mass is a conserved quantity, 3 3 4 4 3 3 R N r π ρ = × π ρ R = r × N 1/3 or R r = N 1/3 ∴ C c = N 1/3 or C = c × N 1/3 Now and Q q V v C c = = ∴ 1/3 1 = ( ) × V Q c N v C C N | | | | | | = × | | | \ . \ . \ . = N 2/3 or V = v × N 2/3 15. An infinite identical capacitors each of capacitance 1 µF are connected as shown in Fig. 5.72. What is the equivalent capacitance between terminals A and B ? [2 µ µµ µµF] Hint. It is clear from the figure that the rows of capacitors are connected in parallel. The capacitance of first row, second row, third row, fourth row ... . is C, C/2, C/4, C/8 ... . The equivalent capacitance of the arrangement is 1 1 1 .... 1 ... 2 4 8 2 4 8 AB C C C C C C | | = + + + + = + + + + | \ . 1 2 2 1 2 F 1 1/ 2 C C = = = × = µ − 16. A spherical capacitor has 10 cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 5. Find the capacitance when (i) the outer sphere is earthed and (ii) inner sphere is earthed. [(i) 3.33 × 10 –10 F (ii) 3.46 × 10 –10 F] Hint. (i) When the outer sphere is earthed, the capacitance of the spherical capacitor is 0 4 ab C K b a = πε − Here 0 9 1 4 9 10 πε = × = C 2 N –1 m –2 ; K = 5; a = 0.1 m; b = 0.12 m ∴ 10 9 1 0.1 0.12 5 3.33 10 F 0.12 0.1 9 10 C − × = × × = × − × (ii) When the inner sphere is earthed, the system is equivalent to capacitors in parallel. One capacitor is between outer sphere and earth and the other capacitor is between outer sphere and inner earthed sphere. The equivalent capacitance is C = 4πε 0 b + 4πε 0 K ab b a − 9 9 1 1 0.1 0.12 0.12 0.12 0.1 9 10 9 10 × = × + × − × × = 0.13 × 10 –10 + 3.33 × 10 –10 = 3.46 × 10 –10 F 17. Find the charge on 5 µF capacitor in the circuit shown in Fig. 5.73. [9 µ µµ µµC] Hint. The p.d. between AB is 6 V. Considering the branch AB, the capacitors 2 µF and 5 µF are in parallel and their equivalent capacitance = 2 + 5 = 7 µF. The branch AB then has 7 µF and Fig. 5. 72 C C C C C C C C C C C C C A B C C C C 16 capacitors Fig. 5. 73 A B 3 F µ 2 F µ 5 F µ 4 F µ 6 V CAPACITANCE 17 3 µF is series. Therefore, the effective capacitance of branch AB is 7 3 21 F 7 3 10 AB C × = = µ + Total charge in branch AB, q = C AB V = 21 10 × 6 = 63 5 µC P.D. across 3 µF capacitor 63 1 21 volts 3 5 3 5 q = = × = ∴ P.D. across parallel combination 21 9 6 volts 5 5 = − = Charge on 5 µF capacitor = (5 × 10 –6 ) × 9 5 = 9 × 10 –6 C = 9 µC 18. Two parallel plate capacitors A and B having capacitance of 1 µF and 5 µF are charged separately to the same potential of 100 V. Now positive plate of A is connected to the negative plate of B and the negative plate of A is connected to the positive of B. Find the final charge on each capacitor. [On A = 200/3 µ µµ µµC; On B = 1000/3 µ µµ µµC] Hint. Initial charge on A, q 1 = C 1 V = (1 × 10 –6 ) × 100 = 100 µC Initial charge on B, q 2 = C 2 V = (5 × 10 –6 ) × 100 = 500 µC When the oppositely charged plates of A and B are connected together, the net charge is q = q 2 – q 1 = 500 – 100 = 400 µC 6 6 Net charge 400 10 200 Final potential difference = V Net capacitance 3 (1 5)10 − − × = = + Final charge on A = C 1 × 200 3 = (1 × 10 –6 ) × 200 3 = 200 3 µC Final charge on B = C 2 × 200 3 = (5 × 10 –6 ) × 200 3 = 1000 3 µC 19. The radii of the two spherical shells which form an air filled spherical capacitor are 10 cm and 10.5 cm. After the inner shell has been charged to a potential of 100 V, the outer shell is taken apart and removed. What is the final potential of the charged inner shell ? [2100 V] Hint. The capacitance of air-filled spherical capacitor is 0 4 ab C b a = πε − Here 4πε 0 = 9 1 9 10 × = C 2 N –1 m –2 ; a = 0.1 m; b = 0.105 m ∴ 9 9 1 0.1 0.105 2.1 F 0.105 0.1 9 10 9 10 C × = × = − × × Charge on inner sphere, 8 9 2.1 21 100 10 C 9 9 10 q CV − = = × = × × When the outer shell is removed, then charge on the inner sphere remains the same. However, the capacitance will change and is given by 8 0 9 1 1 4 0.1 10 F 9 9 10 C a − ′ = πε = × = × × ∴ Final potential of the inner charged shell is 8 10 21 9 10 2100V 9 10 q V C − − ′ = = × × = ′ 20. A capacitor is filled with two dielectrics of the same dimensions but of dielectric constants K 1 and K 2 respectively. Find the capacitances in two possible arrangements [M.N.R.]. 0 0 1 2 1 2 1 2 2 ( ) ( ) ( + ) + 2 A A K K i ii K K d K K d ε ε 18 CAPACITANCE d A B d/2 d/2 K 1 K 2 () i d A B K 1 K 2 ()ii A/2 A/2 Fig. 5. 74 Hint. The two possible arrangements are shown in Fig. 5.74. (i) The arrangement shown in Fig. 5.74 (i) is equivalent to two capacitors in series,each with plate area A and plate separation d/2 i.e., 1 0 1 0 1 2 / 2 K A K A C d d ε ε = = 2 0 2 0 2 2 / 2 K A K A C d d ε ε = = The equivalent capacitance C′ is given by; 1 2 1 0 2 0 0 1 2 1 1 1 1 1 2 2 2 d d d C C C K A K A A K K | | = + = + = + | ′ ε ε ε \ . 1 2 0 1 2 2 K K d A K K | | + = | ε \ . ∴ 0 1 2 1 2 2 A K K C d K K | | ε ′ = | + \ . (ii)The arrangement shown in Fig. 5.74 (ii) is equivalent to two capacitors inparallel,each with plate area A/2 and plate separation d i.e., 1 0 1 0 1 ( / 2) 2 K A K A C d d ε ε = = 2 0 2 0 2 ( / 2) 2 K A K A C d d ε ε = = The equivalent capacitance C ″ is given by; 1 0 2 0 0 1 2 1 2 ( ) 2 2 2 K A K A A C C C K K d d d ε ε ε ′′ = + = + = + ∴ 0 1 2 ( ) 2 A C K K d ε ′′ = + CAPACITANCE 1 PROBLEMS FOR PRACTICE 1. Calculate the radius of a spherical conductor of capacitance 1 farad. [9 × 10 6 km] 2. Can a metal sphere of radius 1 cm hold a charge of 1C ? [No] 3. Calculate the capacitance of a conducting sphere of radius 10 cm situated in air. How much charge is required to raise it to a potential of 1000 V ? [11 pF; 1.1 × 10 –8 C] 4. A capacitor consisting of two parallel plates 0.5 mm apart and each of effective area 500 cm 2 is connected to a 100 V battery. Calculate (i) capacitance of the capacitor and (ii) charge on each plate. [(i) 885 µ µµ µµF, (ii) 0.0885 µ µµ µµC] 5. Find the length of the paper used in a parallel plate capacitor of capacitance 2µF if the dielectric constant of the paper is 2.5 and its width and thickness are 50 mm and 0.05 mm respectively. [Hint. 0 . KA C d ε = Here d = 0.05 mm. Now area, A = length × width] 6. The plates of a parallel plate capacitor are separated by a distance of 0.5 cm. What must be the plate area if the capacitance of the capacitor is to be 2F? [1130 km 2 ] 7. A parallel plate capacitor has circular plates of 8.2 cm radius and 1.3 mm separation. What is the capacitance of the capacitor. If a p.d. of 120 V is applied across the plates, what charge will appear on the plates? [140 pF; 17 nC] 8. A parallel plate capacitor has plates 0.15 mm apart, a plate area of 0.1m 2 and a dielectric of relative permittivity 3. If charge on capacitor plates is 0.5 µC, find (i) electric flux density, (ii) p.d. between plates, and (iii) electric field intensity. [(i) 0.5 × 10 –5 C/m 2 (ii) 28 V (iii) 186667 V/m] 9. Two capacitors of capacitance 2 µF and 4 µF respectively are connected in series. A d.c. potential difference of 900V is applied to this series combination. Find p.d. across each capacitor. [600V; 300V] 10. How can three capacitors of capacitance 3µF, 6µF and 9µF respectively be connected to have a capacitance of 11µF? [3µ µµ µµF and 6µ µµ µµF in series with 9µ µµ µµF in parallel with both] 11. Three capacitors of 2µF, 3µF and 6µF respectively are connected in series across 500V d.c. supply. Find charge and p.d. on each capacitor. [500 µ µµ µµC; 250V, 166.7V, 83.3V] 12. Calculate the capacitance between points A and B in Fig. 5.23. [3µ µµ µµF] 13. In the circuit shown in Fig. 5.24, find the equivalent capacitance between points A and B. [4µ µµ µµF] Fig. 5. 24 Fig. 5. 23 2 F µ 2 F µ 6 F µ 6 F µ 5 F µ 5 F µ 4 F µ 4 F µ 4 F µ 4 F µ A B A B 14. In the circuit shown in Fig. 5.25, find the equivalent capacitance between points A and B.[ 5µ µµ µµF] 15. In the circuit shown in Fig. 5.26, find the equivalent capacitance between points A and B. [1µ µµ µµF] Fig. 5. 26 Fig. 5. 25 2 F µ 4 F µ 3 F µ 2 F µ 3 F µ 2 F µ 3 F µ 3 F µ 3 F µ 3 F µ 3 F µ 1 F µ 4 F µ 1 F µ 8 F µ A B A B 16. How will you combine four capacitors, each of 1µF, to have a net capacitance of 0.75µF? [Three in parallel and fourth in series with this parallel combination] 17. A variable air capacitor has 11 movable plates and 12 stationary plates. The area of each plate is 0.0015 m 2 and separation between opposite plates is 0.001m. Determine the maximum capacitance of this variable capacitor. [292 pF] 2 CAPACITANCE 18. Calculate the number of sheets of tin foil and mica for a capacitor of 0.33 µF capacitance if area of each sheet of tin foil is 82 cm 2 , the mica sheets are 0.2 mm thick and have relative permittivity 5. [183 sheets of tin foil; 182 sheets of mica] 19. A 600 pF capacitor is charged by a 200Vsupply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?. [6 × 10 –6 J] 20. An uncharged capacitor is connected to a battery. Show that half the energy supplied by the battery is lost as heat while charging the capacitor. 21. Two capacitors are connected in parallel and the energy stored is 18J when a potential difference of 6000 V is applied across the combination. When the same capacitors are connected in series, the stored energy is 4J for the same potential difference. What are the individual capacitances ? 2 1 F ; F 3 3 µ µ 22. The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4mm. It is charged to 200V and then the charging battery is removed. Now a dielectric slab (K = 4) of thickness 2mm is placed between the plates. Calculate (i) final charge on each plate (ii) final potential difference between plates (iii) final energy in the capacitor. [(i) 10 –8 C (ii) 125 V (iii) 6.25 × 10 –7 J ] 23. An ebonite plate (K = 3), 6 mm thick is introduced between the parallel plates of a capacitor of plate area 2 × 10 –2 m 2 and plate separation 0.01m. Find the capacitance. [29.5 pF] 24. A parallel-plate capacitor having plate area 100 cm 2 and separation 1.0 mm holds a charge of 0.12 µC when connected to a 120 V battery. Find the dielectric constant of the material filling the gap. [11.3] 25. A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness 3d/4 where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates ? 26. Two parallel-plate capacitors, each of capacitance 40 µF, are connected in series. The space between the plates of the capacitors is filled with a dielectric material of dielectric constant of 4. Find the equivalent capacitance of the system. [32 µ µµ µµF] LONG/SHORT ANSWER QUESTIONS 1. Write a short note on conductors and insulators. 2. Discuss the behaviour of metallic conductors in electric field. 3. What is a capacitor? How does a capacitor store charge? 4. Define the SI unit of capacitance. 5. Derive an expression for the capacitance of an isolated spherical conductor. 6. Derive an expression for the capacitance of a spherical capacitor. 7. What is a parallel plate capacitor? Derive an expression for its capacitance. 8. What do you mean by dielectric constant of a dielectric? 9. Three capacitors are connected in (i) series (ii) parallel. Obtain an expression for their equivalent capacitance. 10. Write a short note on multiplate capacitor. 11. Derive an expression for the capacitance of a cylindrical capacitor. 12. Show that energy stored by a capacitor is q 2 /2C where q is the final charge on the capacitor and C is its capacitance. 13. What do you mean by energy density of electric field? 14. Two capacitors of capacitances C 1 and C 2 are charged to potentials V 1 and V 2 respectively. The capacitors are joined through a conducting wire. What is the value of common potential? 15. Write a short note on polar and non-polar dielectrics. 16. Discuss the behaviour of a dielectric in a uniform electric field. 17. What do you mean by dielectric strength of a dielectric? 0 4 = + 3 K C C K CAPACITANCE 3 18. Derive an expression for the capacitance of a parallel plate capacitor with conducting slab between the plates; assuming that thickness of slab is less than the plate separation. 19. A parallel plate air capacitor has plate separation d. If a dielectric slab of thickness t (t < d) is introduced between the plates, derive an expression for the capacitance of the resulting capacitor. 20. Write a short note on atmospheric electricity. 21. Give an account of thunderstorms and lightning. 22. Write a short note on the discharging effect of sharp points. 23. Write a short note on lightning conductor. 24. Describe the principle, construction and working of Van de Graaff generator. VERY SHORT ANSWER QUESTIONS 1. Why is an insulator called a dielectric? 2. How will you prove that electric field inside a charged conductor is zero? 3. Show that farad is a too large unit of capacitance. 4. Show that SI unit of ε 0 is F/m. 5. Mention three advantages of dielectrics in capacitors. 6. Can you apply any value of electric field to a dielectric? 7. What do you mean by voltage rating of a capacitor? 8. Why are capacitors connected in series or parallel? 9. What is the need of a multiplate capacitor? 10. Name the practical example of a cylindrical capacitor. 11. Show that electrostatic energy of a capacitor is stored in its electric field. 12. What is the importance of energy density of electric field? 13. How will you discharge a charged capacitor? 14. What is the difference between polar and non-polar dielectrics? 15. A parallel plate air capacitor has a capacitance of 10 µF. If air is replaced by mica (K = 6), what is the capacitance of the capacitor? 16. What is the dielectric constant of a metal? 17. What is the importance of capacitance? 18. How does lightning conductor prevent the building from lightning? 19. Why does charge leak off rapidly from pointed ends of a charged conductor? 20. What is the breakdown voltage of air? 2 CAPACITANCE Ans. (i) Conductors have a large number of free electrons while dielectrics have practically no free electrons. (ii) When a conductor is placed in an external electric field, there is no electric field inside the conductor. However, when a dielectric is placed in an electric field, its molecules are polarised. The effect of this polarisation is to weaken the applied electric field within the dielectric. (iii) The dielectric constant of conductors is infinity while that of dielectrics is finite. (iv) The dielectric strength of conductors is zero while that of dielectrics is finite. (v) There is no limit to the current that a conductor can carry, provided that it can be kept cool enough. However, there is a limit to the electric flux that a dielectric will carry without breaking down. Q. 12. Show that capacitance of a metal plate A can be increased by bringing another metal plate B near A. + + + + + + + A + + + + + + + B – + – + – + – + – + – + – + Ans. The charge holding property of a conductor is called its capacitance. Fig. 5.47 shows an insulated metal plate A. Let positive charge be given to it till its potential becomes maximum. No further charge can be given to the plate as it would leak out. Now bring another insulated metal plate B near the plate A as shown in Fig. 5.47. The plate A will induce negative charge on the inner face of B and equal positive charge on the outer face. The induced positive charge tends to increase the potential of plate A while induced negative charge tends to decrease the potenFig. 5.47 tial of plate A. Since the induced negative charge is nearer to plate A than the B A induced positive charge, the net effect is that the potential of plate A decreases. – Therefore, more charge can be given to plate A to raise its potential to maximum + + – value. Thus the capacitance of conductor A is increased by bringing another + + + + – uncharged conductor B near it. Q. 13. In the above question, show that capacitance of metal plate A can be further increased by earthing the plate B. + + + + + + + + – – – – Ans. If the plate B is earthed [See Fig. 5.48], the induced positive charge being free will flow to earth. However, induced negative charge remains since it is bound to the positive charge on plate A. As a result, the potential of plate A is sufficiently Fig. 5.48 reduced. We can now give a large amount of charge to plate A to raise its potential to maximum value. We arrive at a very important conclusion that capacitance of an insulated conductor is increased by bringing near it an uncharged earthed conductor. Q. 14. Two identical metal plates are given charges q1 and q2 (q2 < q1) respectively. If they are now brought close to form a parallel-plate capacitor with capacitance C, what will be the potential difference between the plates ? Ans. Suppose A is the area of each plate. When the plates are held at a distance d, the capacitance of the parallel plate capacitor formed is ε A C= 0 d If E1 and E2 are the electric fields due to two plates, then net electric field E between the two plates is q / A q2 / A 1 (q1 − q2 ) E = E1 − E2 = 1 − = 2 ε0 2 ε0 2 ε0 A ∴ P.D. between plates, V = E d = = 1 d (q1 − q2 ) × d = (q1 − q2 ) 2 ε0 A 2 ε0 A ε0 A   ∵ C = d    Q. 15. A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m2. Will he get electric shock if he touches the metal sheet next morning ? Ans. Yes, the man gets shock. The discharging current in the atmosphere will charge the aluminium sheet. When the man touches the sheet, the charge will flow to earth via the body of the man. Therefore, the man will get a shock. Q. 16. Capacitors P, Q and R have each a capacitance of C. A battery can charge the capacitor P to a potential difference of V. If after charging P, the battery is disconnected from it and the charged capacitor P is connected in the following separate instances to Q and R (i) to Q in parallel and (ii) to R in series, then what will be potential differences between the plates of P in the two instances ? q1 − q2 2C Q. The discharging current due to lightning passes to earth through the metallic body of the car. Since the body of the car is a metal. Ans. all the four capacitors have the same capacitance C. 5. When the capacitor P is charged to a potential difference V. What is the magnitude and nature of charge on plates 1 and 4 ? 12 3 4 5 – V + Ans.e. it acquires charge q = CV.D.CAPACITANCE Ans. Comment. 5. The safest way to protect yourself from lightning is to be inside a car. We know that electric field inside a conductor is zero. Five identical capacitor plates each of area A are arranged such that the adjacent plates are at a distance d apart. The plates are connected to a battery of V volts as shown in Fig. 17.18. ∴ ε0 A V d The plate 4 acts as a negative plate of both the capacitors C3 (between plates 3 and 4) and C4 (between plates 4 and 5). Charge on plate 1 = + C V = ∴ Charge on plate 4 = − (C + C ) V = − 2 ε 0 A V d . the electric field inside it is zero. The system constitutes 4 capacitors in parallel across V volts. Hence potential difference across the plates of P remains V. the circuit is incomplete and no charge is transferred from capacitor P. Q. The plate 1 acts as a positive plate of capacitor C1 (of capacitance C ) formed between plates 1 and 2.49. across P = V/2 V′ = 3 i.49 Since area of each plate is A and the separation between adjacent plates is d. (ii) When P is connected to R in series When P is connected to R in series. (i) When P is connected in parallel to Q Total capacitance = C + C = 2C Total charge =q+0=q ∴ Potential difference V′ across each capacitor is q CV V = = 2C 2C 2 P. Fig. Twenty seven spherical drops. 4πε0 = 9 × 109 ∴ C= 6.09 m. rA= 10 cm .1 − 0. V = Solution: Capacitance of the spherical capacitor is 4π ε0 rA rB C= (rB − rA ) Here rB = 10 cm = 0. (ii) How much charge should be placed on it to raise its potential to 104V? Solution: (i) Capacitance of sphere.1 m.1 1 9 × 109 = 100 × 10–12 F = 100 pF 9 × 109 (0. Assume there is air in the space between the spheres.2/2 = 0. Find the radii of its surfaces. Think about this ! Example 2. What is the capacitance of the capacitor? ∴ Potential of bigger drop. the charge on the bigger drop is 27 × 10–12 C.09 × 0. rB = 12cm .4 × 106 9 × 10 9 = 0. The outer sphere is earthed. r = 6400 km = 6. each of radius 3mm and carrying 10–12C of charge are combined to form a single drop. 9 × 10 Since charge is conserved. (i) Calculate its radius. q = CV = (50 × 10–12) × 104 = 5 × 10–7 C = 0. An isolated sphere has a capacitance of 50pF. we choose earth as a level of zero potential for practical purposes. Find the capacitance and potential of the bigger drop. Solution. Volume of bigger drop = 27 × Volume of smaller drop 4 4 π R 3 = 27 × π r 3 3 3 R = 3r = 3 × 3 = 9 mm = 9 × 10–3 m C = 4 π ε0 R = 1 9 or or Capacitance of bigger drop.CAPACITANCE 1 SOLVED EXAMPLE Example 1: Assuming earth to be an isolated conducting sphere of radius 6400 km. 4π ε0 = ∴ C= 0. r = × C = (9 × 109 ) × (50 × 10−12 ) = 45 × 10−2 m = 45cm 4 πε0 (ii) Charge to be placed. Solution: Given : Here ∴ 4πε0 rA rB = 4π ε0 R rB − rA rB – rA = 2 cm ∴ rA rB =R rB − rA and R = 1. Note: Since capacitance of earth is quite large. C = 4 π ε0 r 1 ∴ Radius of sphere.6m = 60 cm rA rB or rA rB = 120 cm = 60 2 Now (rB + rA)2 = (rB – rA)2 + 4 rA rB = (2)2 + 4 × 120 = 484 ∴ rB + rA = 484 = 22 cm Since rB – rA = 2cm and rB + rA= 22cm.5 µC Example 3.09) Example 5. The thickness of air layer between two coatings of a spherical capacitor is 2 cm. × 9 × 10−3 = 10−12 F = 1pF q 27 × 10−12 = = 27 V C 10−12 Example 4: A spherical capacitor has an inner sphere of radius 9 cm and an outer sphere of radius 10cm. what is the capacitance of earth? Solution: Capacitance of earth. rA = 9 cm = 0. The capacitor has the same capacitance as the capacitance of sphere of 1.2m diameter.4 × 106 m Here. C = 4π ε0 r 1 .711 × 10–3 F = 711 µF This shows that farad is a very large unit of capacitance. Let r and R be the radii of smaller and bigger drops respectively. V = 10 kV = 10 × 103 volts (i) Charge on each plate. This shows that farad is a very large unit of capacitance. Calculate (i) charge on each plate.2 × 0. E = 10% of 107 = 106 Vm–1. d = Solution: Capacitance. What minimum area of the plate is required to have a capacitance of 50 pF? σ ε0 K ∴ Relative permittivity.18 × 10–3 m = 3. Solution: C =300 pF = 300 × 10–12 F.e. K = Solution. of 10 kV is applied to the terminals of a capacitor consisting of two parallel plates. For reasons of safety.5 Example 8. Example 7: What distance apart should the two plates each of area 0.9 × 10-3 m 2 ∴ Plate area A = (50 × 10−12 ) × 10−3 8.2 CAPACITANCE Example 6.797 × 10–6)1000 = 0.85 × 10−12 × 3 . Now E= V d ∴ d = V 1 × 103 = = 10−3 m 6 d 10 = 1. A p.854 × 10−12 Note the enormous magnitude of plate area required to have a capacitance of 1F.854 × 10−12 × 107 Example 10. each having an area of 0.1 × 108 m2 ε0 8. Calculate the capacitance of a parallel plate air capacitor of plate area 30 m2. q = CV = (300 × 10–12) (10 × 103) = 3 × 10–6 C (ii) Electric flux density.797 × 10–3 C = 0.5m? Solution: Area of plate. (iii) potential gradient.1 = 0. The plates of a parallel plate air capacitor are separated by a distance of 1 mm.1 m of a parallel plate air capacitor be placed in order to have the same capacitance as a spherical conductor of radius 0. E = (iv) Electric intensity. ∴ or ε0 A = 4πε0r d A 0. C= ε0 A d d = 1 mm = 10–3 m. Here ∴ A = ?. V = E × d = 500 × 2 = 1000 volts Charge on each plate.797 µF d 2 × 10−3 P.797 mC Example 9. σ = (iii) Potential gradient. r = 0. The resulting capacitance of the arrangement is 300pF.39 ε0 E 8. E = q 3 × 103 = = 3 × 10 –4 C/m 2 A 0. the plates being separated by a dielectric 2 mm thick and of relative permittivity 6. across plates.854 × 10−12 ) (6) (30) = = 0.5 m For parallel plate capacitor. What must be the plate area if the capacitance of the capacitor is to be 1F? Solution: The capacitance of a parallel plate air capacitor is given by.01 m2 separated by a dielectric 1 mm thick. (ii) electric flux density. C = ∴ ε0 K A (8. the electric field E between the plates should not exceed 10% of the dielectric strength of the dielectric i. C = 4π ε0 r A= Since the capacitance of the two capacitors is the same. calculate the charge on each plate.02 = = 3.2 m × 0. q = CV = (0.d.D.02 m2 Radius of sphere. A parallel plate capacitor is to be designed with a voltage rating 1kV using a material of dielectric constant 3 and dielectric strength of 107 Vm–1. C = ε0A/d For spherical conductor. and (iv) relative permittivity of the dielectric. A = 0.18 mm 4π r 4π × 0. If the electric field strength between the plates is 500 V/mm.797 × 10–6 F = 0.01 V 10 × 103 = = 107 V/m d 1 × 10−3 σ 3 × 10−4 = = 3. C = 1F Cd 1 × 10−3 = = 1. 3 0. Solution: (i) For maximum capacitance. all the capacitors will have to be connected in parallel.3 µF and 0. Find (i) charge on each capacitor. (i) and (ii).D.. (ii) P. across each capacitor.5 µF. They are first connected to have maximum capacitance and then connected to have minimum capacitance. 5. ∴ Charge on each capacitor. C1 = C2. the total charge is 750 µC. Since the two capacitors have the same charge and potential. Three capacitors have capacitances of 0. q = CS V = (8. Find the ratio of maximum capacitance to minimum capacitance. CP = 0.14 × 10−3 = = 257 V P. CS = C1 C2 15 × 20 = = 8.(i) . In the circuit shown in Fig.2 = 1 µF (ii) For minimum capacitance. The total capacitance of two capacitors is 4µF when connected in series and 18 µF when connected in parallel. V and C2. Find the values of V1. (iii) C1 = 12 µF or 6 µF.5 mm = 0.57 µF C1 + C2 15 + 20 In series connection. Two parallel plate air capacitors have their plate areas 100 cm2 and 500 cm2 respectively. across 20 µF capacitor = C2 20 × 10−6 Example 14. Solution: Let C1 and C2 be the unknown capacitances.5 + 0. Two capacitors of capacitance 15 µF and 20 µF are connected in series to a 600 V d.3 + 0.. Solution: (i) Equivalent capacitance.5 0.e. C1C2 = 72 .25cm 100 ∴ d2 = A2 d1 A1 Here A2 = 500 cm2 .2 3 CS or ∴ CS = 3/31 µF CP 31 = CS 3 Example 13. 1 1 1 1 31 = + + = 0. all the capacitors will have to be connected in series.CAPACITANCE 3 Example 11.5 mm.D. (i) and (iii). (ii) p.. V1 = Charge on C3 = C3V2 = (8 × 10–6) 20 = 160 × 10–6 = 160 µC ∴ Charge on C2 = 750 – 160 = 590 µC . Then. = (18) 2 − 4 × 72 = ± 6 . C2 = 6 µF or 12 µF Example 15.13. supply. 0. ∴ ε0 A1 ε0 A2 = d1 d2 500 × 0.05 = 0.14 × 10−3 = = 342. If they have the same charge and potential and the distance between the plates of the first capacitor is 0. their capacitances (C = q/V) are equal i.14 × 10–3 C q 5.7 V C1 15 × 10−6 q 5.. what is the distance between the plates of the second capacitor? Solution: Let us denote the first capacitor by suffix 1 and second capacitor by suffix 2.. Solution: Voltage across q (750 × 10−6 ) = = 50 V C1 15 × 10−6 Applied voltage..d.c.(ii) when in parallel when in series Now C1 − C2 = (C1 + C2 )2 + 4C1 C2 Solving eqs.05 cm ∴ d2 = Example 12. A1 = 100 cm2 . charge on each capacitor is the same. across 15 µF capacitor = C1 + C2 =18 C1 C2 =4 C1 + C2 Multiplying eqs. Find the capacitance of each capacitor.2 µF respectively. we get. d1 = 0. V = V1 + V2 = 50 + 20 = 70 V C1 .57 × 10–6) × 600 = 5. 5. 5.14 C1 C4 A 100 pF + 300 V _ B C2 200 pF 100 pF C3 200 pF + 300 V Fig. V1 = VBC = 100 V P.14.d. 5. VBC Charge on C1.4 C2 C1 = 15 µF CAPACITANCE C3 = 8 µF V1 + V Fig. V4 = q 2 × 10−8 = = 200 V C4 100 × 10−12 = 300 – 200 = 100 V P.13 _ V2 = 20 V 590 × 10−6 = 29.c. For 300 V d.15. on various Capacitors  200  × 10−12  × 300 = 2 × 10−8 C Total charge. ∴ Capacitance of C2 = C1 100 pF C2 200 pF C3 200 pF C4 100 pF Fig. supply.D. The equivalent capacitance C′ of series-connected capacitors C2 and C3 is C × C3 200 × 200 C′ = 2 = = 100 pF C2 + C3 200 + 200 The equivalent capacitance of parallel combination of C′ (= 100 pF) and C1 is CBC = C′ + C1 = 100 + 100 = 200 pF The entire circuit now reduces to two capacitors C4 and CBC (= 200 pF) in series. The above network can be redrawn as shown in Fig.5 µF 20 Example 16: Obtain the equivalent capacitance for the network shown in Fig.D.D.D. across C3 = 100/2 = 50 V Charge on C2 = Charge on C3 = Total charge – Charge on C1 = (2 × 10–8) – (10–8) = 10–8 C _ . across C1.D. q = CV =  3   ∴ ∴ Charge on C4 = 2 × 10–8 C P. across C4. 5. determine the charge and voltage across each capacitor.15 C Solution: Equivalent Capacitance. across C2 = P.5 × 10–6 F = 29. 5. between B and C. ∴ Equivalent capacitance of the network is C × CBC 100 × 200 200 C= 4 = = pF C4 + CBC 100 + 200 3 Charges and p. q1 = C1 VBC = (100 × 10–12) × 100 = 10–8 C P. 1 2 5 6 3 4 B and plate 5). equivalent capacitance between A and B is CAB = C1 + C2 + C3 Fig. 5. Solution: (i) CAB = 10 µF 8 µF 8 µF 8 µF A 10 µF B 8 µF C 12 µF D 8 µF + 400 V _ Fig.19.18. 5. C1 C2 C3 the three capacitors are in parallel. C1 = 3 µ F A B A C1 B C2 = 9 µ F C4 = 9 µ F C3 = 9 µ F Fig. Similarly. If a 10V battery is connected between A and B. Hence. C3 and C4 is given by. The equivalent capacitance C′ of the series connected capacitors C2.18 Example 19: In the circuit shown in Fig. find (i) the equivalent capacitance between A and D and (ii) the charge on 12 µF capacitor.18. 5. 5.16 C2 C3 Fig.17 C4 Solution: The given network is equivalent to the network shown in Fig. plate 3 and plate 6) is connected to point B. 5. Determine the equivalent capacitance between points A and B.17.16 shows a network of four capacitors.CAPACITANCE 5 Example 17: Fig. 1 1 1 1 1 1 1 1 = + + = + + = C ′ C2 C3 C4 9 9 9 3 ∴ C′ = 3 µF Between points A and B. Therefore. other plate of each capacitor (plate A 2. the equivalent capacitance CAB between points A and B of the network is CAB = C1 + C′ = 3 + 3 = 6 µF Total charge stored on the capacitors is q = CAB × V = (6 × 10–6) × 10 = 60 × 10–6 C = 60 µC Example 18: Calculate the equivalent capacitance between points A and B in Fig. we now have capacitors C1 and C′ (= 3 µF) in parallel. Solution: Refer to Fig. plate 4. how much total charge will be stored on the capacitors. 5. 5. Therefore. 5. It is clear that one plate of each capacitor is connected to point A (plate 1. 5.19 8×8  + (8) + (8) = 4 + 8 + 8 = 20 µF CBC =  8 + 8 C × CBC 10 × 20 20 C AC = AB = = µF C AB + CBC 10 + 20 3 CCD = 8 + 12 = 20 µF ∴ C AD = (20 / 3) × 20 = 5 µF (20 / 3) + 20 . Find the capacitance of the capacitor.6 (ii) Total charge. 5. 5. between C and D. 5.21 (ii) shows the condition when the inner plate is three times as near as one plate as the other. Capacitance of this capacitor.20 (iii). Find the equivalent capacitance between points A and B.002C CAPACITANCE The total charge will divide between the parallel capacitors connected in the branch CD. the equivalent capacitance C′ of the series connected capacitors C1 and C2 is C C 8×8 C′ = 1 2 = = 4 µF C1 + C2 8 + 8 The equivalent capacitance C″ of series connected capacitors C3 and C4 [See Fig. 5. 5.21 (ii) Fig. 5.1062 × 10–6 F = 0. 5.1062 µF Example 22: A parallel plate capacitor has three similar parallel plates. 5. capacitance is given by.20 (i) can be drawn as shown in Fig.2 × 10–3 C = 1. We find that the circuit is a Wheatstone bridge. It means that points C and D are at the same potential.854 × 10−12 ) × 6 × (5 × 10−4 ) 0. C3 C4 8×8 = = 4 µF C1 + C2 8 + 8 CAB = C′ || C″ = 4 || 4 = 4 + 4 = 8 µF C ′′ = Now Example 21: A mica dielectric parallel plate capacitor has 21 plates. there will be no charge on capacitor C5.005 × 10−3 = 0.002 P.20 (iii). ε K A ε 0 K A 4ε 0 K A C1 = 0 + = d /2 d /2 d d/2 d/4 d/2 3d/4 (i) Fig. Find the ratio of capacitance when the inner plate is mid-way between the outers to the capacitance when inner plate is three times as near one plate as the other.21 (i). Since the product of opposite arms of the bridge are equal (C1C4 = C2C3 because C1 = C2 = C3 = C4). Take the relative permittivity of mica as 6. q 0.D. q = CAD × V = (5 × 10–6 ) × 400 = 0. C1 = C2 = C3 = C4 = 8 µF and C5 = 10 µF.20 C C2 C5 C4 B A C1 C C2 B C3 D (iii) C4 Solution: A little reflection shows that circuit of Fig. VCD = = = 100 V CCD 20 × 10−6 ∴ Charge on 12 µF capacitor = (12 × 10–6) × 100 = 1.20 (i). ε KA C = ( n − 1) 0 d = (21 − 1) (8. . shows the condition when the inner plate is mid-way between the outer plates. Solution: Fig.2 µC Example 20: In the network shown in Fig.20 (iii)]. Solution: For a multiplate capacitor. C1 C1 C5 D A C 3 C C2 B A C3 C4 (i) D (ii) Fig. 5. Hence. 5. each having an effective area of 5 cm2 and each separated by a gap of 0.005mm. Referring to Fig. Therefore. the bridge is balanced.20 (ii). The arrangement is equivalent to two capacitors in parallel. this capacitor is ineffective and can be removed from the circuit as shown in Fig. across A is 120V.s across the capacitors are in the inverse ratio of their capacitances. of 250V. between the two cylinders. ε K A ε0 K A 16ε0 K A C2 = 0 + = d /4 3d / 4 3d ∴ C1/C2 = 0.0005 C When the two capacitors are connected through a wire. the potential of the inner cylinder is equal to p. This p. Calculate the new p. Total charge 0. Solution: Capacitance of each capacitor. the total capacitance CP = 5 + 3 = 8 µF. ∴ P. charge on each capacitor is the same.d.d. Solution: Let C1 and C2 µF be the capacitances of the capacitors A and B respectively. across capacitors = . The capacitors are disconnected from the supply and are reconnected in parallel with each other. and charge on each capacitor. supply.0006 C Since the capacitors are connected in series. across 4 µF capacitor = 250 × 6 = 150 V 4+6 Charge on 4 µF capacitor = (4 × 10–6) × 150 = 0. number of such rows in parallel = C/CS = 0.1 µF. is increased to 140V when a 3µF capacitor is connected in parallel with B. 5. ∴ P. Calculate p.D. Calculate the number of capacitors needed if it is desired to obtain a capacitance of 0.D.0012 C is distributed between the capacitors to have a common p. The p. The charge 0. Solution: (i) The capacitance of a cylindrical capacitor is Kl 5 × 0.d. V = 60 V + _ C C C C 7 60 V Fig. When the capacitors are connected in series [See Fig. ∴ Total number of capacitors = 4 × 4 = 16 Fig. Calculate the capacitance of A and B. ∴ Charge on both capacitors = 2 × 0.22 Since each capacitor can withstand 15 V.d.0012 = 120 V = CP 10 × 10−6 Charge on 4 µF capacitor = (4 × 10–6) × 120 = 480 × 10–6 = 480 µC Charge on 6 µF capacitor = (6 × 10–6) × 120 = 720 × 10–6 = 720 µC Example 27: Two capacitors A and B are connected in series across a 200V d. charge on each capacitor is the same.d.1µF for use in a circuit involving 60V. i.1 µF Voltage rating of each capacitor.1/4 = 0. VC = 15 V Supply voltage. determine (i) capacitance of cylindrical capacitor and (ii) potential of inner cylinder.0005 ∴ P. q = CV = (5 × 10–6) × 100 = 0.1/0.31 (i)].025 = 4.d.22 shows the arrangement of capacitors.d. The relative permittivity of the insulation is 5. 5. q 0. Since it is desired to have a total capacitance of 0.4log10 (b / a ) 41. When the capacitors are connected in parallel.c.4 /15) = 2.4 cm and 15 cm respectively.75 Example 23: Given some capacitors of 0.e. potential of inner cylinder = 5000V.d.CAPACITANCE Capacitance C2 of this capacitor. across capacitors = = = 62.11 × 10–9F (ii) Since the outer cylinder is earthed. Example 24: A cylindrical capacitor has two co-axial cylinders of length 20 cm and radii 15.0005 C is distributed between the two capacitors to have a common p.d.. Solution: In series-connected capacitors. C = 0.d.d. Solution: Charge on 5 µF capacitor. The total charge 0. of 100V and then connected to an uncharged 3 µF capacitor.1µF capable of withstanding 15V. 5.D. across the capacitors. of 5000V is maintained between the two cylinders. the total capacitance CP = 4 + 6 = 10 µF. If a p. p.4log10 (15. Example 25: A 5µF capacitor is charged to a p.5 V CP 8 × 10−6 Example 26: Two capacitors of capacitance 4 µF and 6 µF respectively are connected in series across a p.0006 = 0.2 C= × 10−9 F = × 10−9 41. Capacitance of 4 series-connected capacitors.025 µF. the number of capacitors to be connected in series = 60/15 = 4.0012 C Parallel connection. CS = C/4 = 0. If the distance between the plates is halved. After being disconnected. The charge on the capacitor is 1. q= V ×r 9 9 × 10 9 × 10 1 1 = qV = (0. Solution: Potential at the surface of sphere. The charge 1.67 × 10−3 ) × (3 × 106 ) = 1005 J Energy stored in sphere 2 2 When the sphere is earthed.(ii) 3 µF A C1 120 V + _ B C2 80 V A C1 140 V + _ B C2 60 V 200 V (i) Fig. V = Example 29: A metal sphere 4 m in diameter is charged to a potential of 3 MV.6 × 10–3 C distributes between the two capacitors to have a common p. it is immediately connected to an uncharged capacitor of 4 µF. the charge on the capacitor (q′ = C′V)becomes twice. Determine (i) the p. A parallel plate capacitor is connected to a 12 V battery.31 (ii)]. The extra charge is supplied by the battery.064 J 2 2 It may be noted that there is a loss of energy. (i) and (ii). ∴ C1 × 140 = (C2 + 3) 60 or 7C1 – 3C2 = 9 . Therefore. the total capacitance. 5. Thus the circuit shown in Fig. C1 = 16 µF. 9 = q r (3 × 106 ) × 2 = 0. When the plate separation is decreased to half. A parallel plate 100 µF capacitor is charged to 500V.d.31 (ii) can be thought of as a series circuit consisting of capacitances C1 and (C2 + 3) connected in series. of V volts. across capacitors. what will be the new potential difference between the plates and what will be the change in the new stored energy? .35 × 10–10 C.08 J 2 2 ∴ P..6 µF. After joining When the capacitors are connected through a wire.4 µF Example 28: A 16 µF capacitor is charged to 100V. across the combination (ii) the electrostatic energies before and after the capacitors are connected. stored energy will be dissipated as heat in the resistance wire.35 × 10–10 C Example 31. the combined capacitance of this parallel branch is (C2 + 3).5 C1 . Calculate the heat generated when the sphere is earthed through a long resistance wire.. U 2 = CPV 2 = (20 × 10–6) × (80)2 = 0. Cp = C1 + C2 = 16 + 4 = 20 µF. V = 9 × 109 ∴ Charge on sphere. C1 = 3.6 × 10−3 = 80 V = CP 20 × 10−6 1 1 Energy stored.6 × 10–3 C U1 = 1 1 C1 V12 = (16 × 10−6 ) × 1002 = 0. find the extra charge given by the battery.(i) When a 3 µF capacitor is connected in parallel with B [See Fig. Energy stored. C2 = 5. q 1. Solution: Before joining Charge on 16 µF capacitor. Solution. If the plate separation is decreased to half..31 200 V (ii) Solving eqs. C2 = 4 µF q = C1V1 = (16 × 10–6) × 100 = 1. 5.67 × 10−3 C Extra charge supplied by battery = q′ – q = 2q – q = q = 1..8 CAPACITANCE ∴ C1 × 120 = C2 × 80 or C2 = 1.d. 5. we have. the capacitance (C = ε0 A/d) becomes twice. Example 30.D. This is primarily due to the heat dissipated in the conductor connecting the capacitors. Calculate the capacitance of the capacitor. d = 0.01m.5 × 10−3 0. the new capacitance C′ becomes twice i. the charge on the capacitor remains the same..01 m.25 × 10–3 J – Tube In order to produce the flash. The dielectric constant for ebonite is 3. C′ = 2C. find the relative permittivity of the additional dielectric.5 V = 2.32 shows a circuit for a camera flash.95 × 10–12 F Example 34: A parallel plate capacitor has plate area of 2 m2 spaced by three layers of different dielectric materials. −3 = (8. Fig. Solution: Capacitance of capacitor. Solution: Fig. K = 3 C= (8. C= ε0 K1 A ε 0 × 4 × A = d d ε0 A ε0 A C = = −3 t1 t2 3 3 × 10 5 × 10−3 + + K1 K 2 4 K2 . 5. 5.01 − 6 × 10−3 (1 − 1/ 3) = 2.40 (i) and Fig.1 × 10–3 s). C = ε0 A d − t (1 − Here ∴ 1 ) K A = 2 × 10–3 m2. 4.5)2 2 2 + 2000 µF Discharge 1. When a flash is required. Fig.CAPACITANCE 9 Solution: C = 100µF = 100 × 10–6F = 10–4F. Since the capacitor is not connected to the battery.3 mm respectively.5 and 0. ∴ q = CV = C′V′ or V′ = CV CV V 500 = = = = 250 volts C′ 2C 2 2 2 New stored energy = 1 1 V  C ′V ′2 = (2C )   2 2 2 1 CV 2 1  1  = =  CV 2  2 2 22  1 1  =  × 10−4 × (500)2  = 6.5V cell. 5. C = ε0 A t t1 t + 2 + 3 K1 K 2 K3 = 0.(ii) . The potential difference between the plates must decrease to maintain the same charge.1 ms (= 0.. t = 6 × 10–3 m.854 × 10−12 ) × 2 × 10−3 0. the energy stored in the capacitor is made to discharge through a discharge tube in 0.25 × 10−3 = = 22. 1. The relative permittivities are 2.5. Solution: ∴ Power of flash = Capacitance of the capacitor. 6 and thicknesses are 0. The area of the plates of the capacitor is 2 × 10–3 m2 and the distance between them is 0. An additional piece of insulation 5 mm thick is now inserted between the plates.40 (ii) respectively show the two cases.026 × 10–6 F 0. If the capacitor has now capacitance one-third of its original capacitance.854 × 10−12 ) × 2 For the first case.5 × 10 1.32 U 2.1 × 10 Example 33: A parallel plate capacitor is partially filled with an ebonite plate of thickness 6 mm. V = 500 volts When plate separation is decreased to half. Solution: Electronic Energy stored in the capacitor is Trigger 1 1 U = CV 2 = × (2000 × 10−6 ) × (1. A 2000 µF capacitor is charged by 1. For the second case. the capacitor discharges in 0.5 W −3 Time 0.e. Calculate the energy stored in the capacitor and power of the flash. 5.. Calculate the capacitance of the capacitor..3 × 10−3 + + 2 4 6 Example 35: A capacitor is composed of two plates separated by 3 mm dielectric of relative permittivity 4.25 J 2 2  Example 32.1 ms giving a powerful flash.(i) . the capacitance would have been ε0 A . The distance between the plates is effectively reduced by t – (t/K)..d. (ii) p. and (iv) energy loss. C0 = Cm = ε0 A d −t+ t K .10 CAPACITANCE K1 = 4 K1 = 4 K2 = ? 3 mm (i) Fig. In order to bring the capacitance to the original value. Now a dielectric slab (K = 4) of thickness 2 mm is introduced between the plates. ∴ New separation between plates = d + ( t – t /K) = 5 + ( 2 –2/3) = 6. thickness 2 mm and relative permittivity 3 is now introduced between the plates. Solution: When a dielectric slab of thickness t (t < d) is introduced between the plates of air capacitor.(i) d When a dielectric slab of thickness t is placed between the plates of the capacitor. the plates must be separated by this much distance in air. (i) and (ii) reveals that with the introduction of the dielectric slab between the plates of air capacitor. its capacitance increases to Cm and p.. C= ε0 A d − (t − t / K ) . the charge on capacitor plates will remain the same after the introduction of the dielectric slab. capacitance is given by. Final charge on each plate = q = 10–8 C (ii) When dielectric slab is placed between the plates of the capacitor. what must be the new separation between the plates to bring the capacitance to the original value. (i) by eq. (iii) final energy in the capacitor. C0 = ε0 A .33 Example 36: An air capacitor has two parallel plates of 1500 cm2 area and held 5 mm apart. its capacitance increases. between plates decreases to V.d. (ii).. between the plates. It is charged to 200V and the charging battery is removed.... Solution: The capacitance of parallel plate air capacitor is given by. Find (i) final charge on each plate. charge (q) on each plate is q = C0 × V0 = (50 × 10–12 ) × 200 = 10–8 C After Introduction of the Dielectric Slab (i) Since the battery is removed before the introduction of the dielectric slab. the capacitance is given by.. 3 = 4 3 5   +  3  4 K2  ∴ K2 = 3. If a dielectric slab of area 1500 cm2. 5.40 3 mm (ii) 5 mm Dividing eq.33 mm Example 37: The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4 mm.(i) If the medium were totally air.(ii) ∴ C0 d − t + (t / K ) 4 − 2 + (2 / 4) 5 = = = Cm d 4 8 Before Introduction of the Dielectric Slab When air capacitor is charged to 200 V and then battery is removed.(ii) d An inspection of eqs.. q = C0 V0 = Cm V ∴ V = C0 5 V0 =   × 200 = 125 V Cm 8 . e.e.e. Capacitance of capacitor.d.CAPACITANCE Note that p. between the plates decreases. Final stored energy. C′ = KC = 6 × 17. (ii) After the voltage supply is disconnected C′ = KC = 6 × 17. the charge on the plates remains the same.7 pF . Explain what would happen if a 3 mm thick mica sheet of dielectric constant 6 were inserted between the plates (i) while the voltage supply remains connected (ii) after the supply was disconnected.7 = 106. Calculate the minimum radius of the spherical shell. Dielectric constant of mica. Since C = q/V and V is same but C has increased. the energy stored in the capacitor would have been smaller by 1/K i. U = Note: Initial stored energy.77 × 10–9 C Since the supply is disconnected. U = qV 2 2 1 ∴ Energy loss = U0 – U = q (V0 – V) 2 1 = (10–8) (200 – 125) = 3. the capacitance becomes K times i. between the plates of the capacitor remains equal to 100 V.75 × 10–7 J. The dielectric strength of the gas surrounding the electrode is 5 × 107 V/m.d.2 × 10−12 Example 39.2 pF Charge on capacitor. The capacitor is connected to a 100V supply. (iii) Final energy stored in the capacitor.75 × 10–7 J 2 This loss of energy will be apparent to the person who introduced the slab. U0 = qV0 .7 × 10–12 × 100 = 1. C = 17. q = CV = 17. U0/K. the charge on capacitor must increase i.2 × 10–12 × 100 = 1.25 × 10–7 J Had the slab been 4 mm thick (= distance between plates). q′ = C V = 106. Because the capacitance (C = q/V) has increased.2 pF The p. K = 6 (i) When voltage supply remains connected When mica sheet is inserted between the plates of the capacitor.7 = 106.77 × 10−9 = = 16.67 V C ′ 106. Electric potential V of a charged shell is V = 1 q 4 π ε0 r 1 q 4 π ε0 r 2 Electric field at the surface of a charged shell is E= ∴ V =r E or r= V 25 × 105 = = 5 × 10−2 m = 5cm 7 E 5 × 10 . A parallel plate capacitor having plate separation of 3mm possesses a capacitance of 17. V′ = q 1.7 pF. the potential difference across the plates must decrease to maintain the same charge. Example 38. 11 1 1 qV = (10−8 ) × 125 = 6. the shell electrode is at 25 × 105 V. Charge on capacitor. Solution. This is in agreement with theory.06 × 10–8 C The extra charge is supplied by the battery. Thus we arrive at a very important conclusion that final energy after the slab filling entire space is introduced is smaller by 1/K. In a Van de Graaff generator. U0 = 1 1 qV0 = (10–8) × 200 = 10– 6 J 2 2 Final stored energy U = 6. The capacitor would exert a tiny force on the slab and would do work on it equal to 3. Solution.25 × 10 –7 J 2 2 1 1 (iv) Initial stored energy. C.31 µC] Hint.002 = 0.05 × 10–3 m U = ∴ ∴ A= Length = Cd (2 × 10−6 ) × (0.05 mm respectively. A = 6 10–4 m2.7. then equivalent capacitance C′ is C′ = 5 × 20 = 100 µF = 100 × 10–6 F Energy stored is given by.] [15 µF] Hint. (i) Here C= ε0 K A d ε0 = 8.C. The dielectric constant of paper is 3. A capacitor of 20 µF and charged to 500 V is connected in parallel with another capacitor of 10 µF charged to 200 V.012 C Total capacitance. (i) Find the capacitance of the paper capacitor. What is the capacity of the uncharged capacitor ? [E. q = q1 + q2 = 0.T.002 C Total charge on capacitors.7. (ii) What is the maximum charge that can be placed on the capacitor ? The dielectric strength of paper is 16 × 166 V/m.12 CAPACITANCE MORE NUMERICAL PROBLEMS 1. q2 = C2 V2 = (10 × 10–6) × 200 = 0. A parallel-plate capacitor has plates of dimensions 2 cm × 3 cm.01 C Charge on second capacitor.T.5 Area 4. K = 3. V = 3 volts.A.E. d = 0.85 × 10–12 C2 N–1 m–2. A 5 µF capacitor is fully charged across a 12 V battery and connected to an uncharged capacitor.6 × 10−12 F 1 × 10−3 (ii) Since the thickness of the paper is 1 mm.5 m = 90 m = Width 50 × 10 −3 4.85 × 10−12 ) × (3. C2 = ? 3= 5 × 12 + C2 × 0 5 + C2 ∴ C2 = 15µF 5. What is the total energy stored in these when connected in parallel and charged to 400 V ? [E.M. q1 = C1 V1 = (20 × 10–6) × 500 = 0. 1 1 C′ V2 = × (100 × 10–6) × (400)2 = 8 J 2 2 3.5 m 2 −12 ε0 K 8. V2 = 0.M.5 and its width and thickness are 50 mm and 0.05 × 10−3 ) = = 4.012 = = 400 V C 30 × 10−6 2.A. [90 m] ε0 K A Hint. V = Hint. 91] [8 J] ∴ Common potential. The plates are separated by a 1 mm thickness of paper. Five equal capacitors connected in series have a resultant capacitance of 4 µF. C = 4 or C = 4 n = 4 × 5 = 20 µF n When the capacitors are connected in parallel. The common potential V after connection is V = C1 V1 + C2 V2 C1 + C2 Here ∴ C1 = 5 µF.85 × 10 × 2.01 + 0. Since 5 (= n) capacitors are connected in series. the maximum voltage that can be applied before breakdown occurs is C= (8. d = 1 10–3 m = 19. Find the length of the paper used in a capacitor of capacitance 2 µF if the dielectric constant of the paper is 2. Find the common potential. C= d Here C = 2 µF = 2 × 10–6 F.6 × 10–12 F (ii) 0.7) × (6 × 10−4 ) .E. [Roorkee] [400 V] Hint. Charge on one capacitor. Suppose C µF is the capacitance of each capacitor. The voltage across it is found to be 3 V. C = C1 + C2 = (20 × 10–6) + (10 × 10–6) = 30 × 10–6 F q 0.05 mm = 0.5. V1 = 12 volts. K = 2. [(i) 19. 5. the maximum charge that can be placed on the capacitors is 6 × 10–3 C ( = q1).e.. the electric field and the stored energy ? Hint. Find the energy stored in the capacitor before and after the dielec- ∴ Vmax = tric is inserted. the capacitor is kept connected with the battery and then dielectric is inserted between the plates.N. it gets pulled into the device.6 × 10–12) × (16 × 103) = 0. electric field ( = V0/d) will also remain unchanged..70 (ii)].. Hint.31 µC 6. (i) 2 C0 Eq. This work is simply = U0 – U.70 (ii) ∴ Energy stored in the capacitor after insertion of dielectric is q02 q02 U = = 0 2 C ′ 2 K C0 K U or . charge on the capacitor remains the same. the capacitance. What maximum voltage will the system of these two capacitors withstand if they are connected in series ? [M.31 × 10–6 C = 0. But the capacitance of the capacitor is increased K times i. Therefore. the positive work done by the system = U0 – U.70 (i). Energy stored in the capacitor in the absence of dielectric is U0 = 1 C0 V02 2 V0 = q0/C0. A parallel-plate capacitor is charged with a battery to a charge q0 as shown in Fig. 1992] [9 kV] Hint. The external agent must do negative work to keep the dielectric from accelerating. 0 2 C0 2 K C0 Since q02 . when capacitors are connected in series.CAPACITANCE 13 Vmax = Emax × d Here Emax = 16 × 106 V/m.. U0 = q0 + C0 _ q0 + KC0 _ V0 (i) Fig. How will you account for “missing energy” ? When the dielectric is inserted into the capacitor. . As a result. Since the battery remains connected. 5. (i) gives the energy stored in the capacitor in the absence of dielectric. (ii) U = 0 K Since K > 1. Suppose in the above problem. the potential difference V0 will remain unchanged. we find that final energy is less than the initial energy by the factor 1/K. d = 1 mm = 1 × 10–3 m ∴ Maximum charge that can be placed on capacitor is qmax = C Vmax = (19. q1 q 6 × 10−3 6 × 10−3 + 1 = + C1 C2 1 × 10−6 2 × 10−6 = 6 × 103 + 3 × 103 = 103 (6 + 3) = 9 × 103 V = 9 kV 7. Alternately.R. this can be expressed as : q0 2 q2 . the potential difference. After the battery is removed and the dielectric is inserted between the plates. new capacitance is C′ = K C0 [See Fig 5. The battery is then removed and the space between the plates is filled with a dielectric of dielectric constant K. What will be the change in charge. U = 8. The maximum charge q1 and q2 that can be placed on C1 and C2 are q1 = C1V1 = (1 × 10–6) × (6 × 103) = 6 × 10–3 C q2 = C2V2 = (2 × 10–6) × (4 × 103) = 8 × 10–3 C The charge on capacitor C1 should not exceed 6 × 10–3 C. A capacitor of capacitance C1 = 1 µ F withstands the maximum voltage V1 = 6 kV while another capacitance C 2 = 2 µ F withstands the maximum voltage V 2 = 4 kV.. In this case. When a 3mm slab is introduced between the plates. charge q remains the same. the capacitance of the capacitor is C1 = 50 pF = 50 × 10–12 F With dial at 180°. the capacitor is disconnected from the battery and the dial is turned at 0°.] [(i) 7600 V (ii) 1. the capacitance of the capacitor is C2 = 950 pF = 950 × 10–12 F P. With dial set at 180°. V2 = 400 V ∴ Charge on C2. the capacitor is connected to 400 V battery.4 × 10–3 – 3 × 10–3  1 −  K  1  or 2.R.] [K = 5] Hint. q = C V0 = KC0 V0 = K q0 CAPACITANCE 1 C0 V02 2 1 1 Final stored energy.. t = 3 mm = 3 × 10–3 m 1  ∴ d = d + 2. 9. A parallel plate capacitor is maintained at a certain potential difference.14 The capacitance C0 will increase to C = K C0 The charge will also increase to q = K q0 as explained below. Will any work be done in inserting the dielectric? The answer is yes. The battery not only gives the increased energy to the capacitor but also provides the necessary energy for inserting the dielectric.. Find the dielectric constant of the slab. Suppose V1 is the potential difference across the capacitor when the dial reads 0°. (i) With dial at 0°.R. [M.4 × 10–3 = 3 × 10–3  1 −  K  ∴ K=5 10. (ii) ∴ or Here ε0 A ε0 A = d d ′ − t (1 − 1/ K ) d = d′ – t (1 – 1/K ) d ′ = d + 2.. (i) What is the potential difference across the capacitor when the dial reads 0° ? (ii) How much work is required to turn the dial if friction is neglected ? [M. C′ = .N. the distance between the plates is increased by 2.D.4 × 10–3 m .. (i) d With the introduction of slab of thickness t. ∴ q = C1V1 q 380 × 10−9 = = 7600 V C1 50 × 10−12 (ii) Work required = Gain in energy of the capacitor or V1 = 1 1 C1 V12 − C2 V22 2 2 1 1 = × 50 × 10−12 × (7600) 2 − × 950 × 10−12 × (400)2 2 2 = 1. q0 = C0 V0 . U 0 = C0 V02 = K C0 V02 = KU 0 2 2 ∴ U = KU0 Note that stored energy is increased K times. U 0 = ε0 A d ′ − t (1 − 1/ K ) Now the charge (q = CV) remains the same in the two cases. The capacitance of parallel-plate capacitor in air is ε A C= 0 .N. in order to maintain the same potential difference. the new capacitance is Initial stored energy. across C2.37 × 10–3 J] Hint. The capacitance of a variable radio capacitor can be changed from 50 pF to 950 pF by turning the dial from 0° to 180°. the work will be done by the battery.37 × 10–3 J = . After charging. q = C2V2 = (950 × 10–12) × 400 = 380 × 10–9 C When the battery is disconnected.4 mm. (ii) What is the potential of the inner sphere ? (iii) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. 5.13 − 0.D. . N drops of mercury of equal radii and possessing equal charges combine to form a big drop.CAPACITANCE 15 11.33 × 10−11 F 9 × 109 Note that capacitance of an isolated sphere is much smaller than that of the concentric spheres.5 µC. Total charge. across series combination.6 × 10 12.5 × 10−6 = = 4. one plate of C3 being earthed. K = 32.12 14. three capacitors being connected in series ? [20 V] Hint. What is the charge. potential and capacitance of the individual small drop.12 (ii) Potential of inner sphere is q 2. The equivalent capacitance of this series combination is given by. V and C. The outer sphere is earthed and the inner sphere is given a charge of 2. q = CV = (90 × 10 Now q = ne ∴ Number of electrons transferred is q 1. 1 1 1 1 1 1 1 3 = + + = + + = C C1 C2 C3 20 30 15 20 20 20 µF = × 10 −6 F 3 3 P. a spherical capacitor has large capacitance. v and c be the charge.V2 = = = 20 V C2 30 × 10−6 13. a = 0. the total potential is distributed over two spheres and the potential difference between the two spheres becomes smaller. Since capacitance (C = q/V) is inversely proportional to potential difference. = 4πε0 r = 0. A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. across C2 . b = 0. If C1 = 20 µF. Why is the capacitance smaller in the latter case ? [(i) 5. (i) Capacitance of spherical capacitor is given by.55 × 10−9 (iii) Capacitance of isolated sphere is V = = 1.5 × 102 V C 5.1 × 10−9 n= = = 6.55 × 10–9 F (ii) 4. A 90 pF capacitor is connected to a 12 V battery and is charged to 12 V. The space between the concentric spheres is filled with a liquid of dielectric constant 32. It is because in case of concentric spheres. Let q.13 m. ab C = 4πε0 K b−a Here ∴ 1 = 9 × 109 C2N–1 m–2.13 × = 5.1 × 10–9 C Hint.D. The corresponding quantities for the bigger drop are Q.33 × 10–11 F] Hint. q = CV = (i) Determine the capacitance of the capacitor.9 × 109 −19 e 1. C2 = 30 µF and C3 = 15 µF and the insulated plate of C1 be at a potential of 90 V.12 × 0.55 × 10 −9 F 9 × 109 0.5 × 102 V (iii) 1. Charge on bigger drop = N × charge on small drop ∴ Q = Nq The capacitance of a spherical drop is proportional to the radius. what is the potential difference between the plates of C2. Fig. How many electrons are transferred from one plate to the other ? [6.12 m 4πε0 C= 32 0. V = 90 – 0 = 90 V C= 20 × 10–6 ×90 = 6 × 10–4 C 3 q 6 × 10−4 ∴ P.71 shows the conditions of the problem. capacitance and potential of the bigger drop ? [(i) Q = Nq (ii) C = c N1/3 (iii) V = v N2/3] Hint. one plate of C3 potential of 90 V.9 × 109] –12) × (12) = 1. Fig. . the capacitance of the spherical capacitor is ab C = 4πε0 K b−a 1 Here 4πε0 = = C2 N–1 m–2.. = C 1 + + + + .. Considering the branch AB.. 5.72.d. It is clear from the figure that the rows of capaciC 16 capacitors tors are connected in parallel. An infinite identical capacitors each of capaciC C C tance 1 µF are connected as shown in Fig. (i) When the outer sphere is earthed. fourth row .16 C R = c r Since mass is a conserved quantity.1 × 0. the capacitors 2 µF and 5 µF are in parallel and their equivalent capacitance = 2 + 5 = 7 µF. The space between the two is filled with a dielectric of dielectric constant 5.12 − 0. K = 5.12 + × 3 µF 9 × 109 9 × 109 0. 5. the system is equivalent to capacitors in parallel. A spherical capacitor has 10 cm and 12 cm as the radii of inner and outer spheres.1 = 0. Find the capacitance when (i) the outer sphere is earthed and (ii) inner sphere is earthed.33 × 10 −10 F 9 0.. 5. third row. Find the charge on 5 µF capacitor in the circuit shown in Fig.33 × 10–10 F (ii) 3.1 m. [(i) 3.73 B ..  2 4 8 2 4 8    1  = C  = 2C = 2 × 1 = 2 µF 1 − 1/ 2  16.12 − 0. b = 0..72 The equivalent capacitance of the arrangement is C C C 1 1 1   C AB = C + + + + . C/4.. A C/2. The equivalent capacitance is ab C = 4πε0 b + 4πε0 K b − a 2 µF 1 1 0.73.13 × 10–10 + 3.33 × 10–10 = 3..1 9 × 10 (ii) When the inner sphere is earthed.12 ∴ C= ×5× = 3. The p. second row.12 = × 0. . What is the equivalent capacitance between terC C C C C minals A and B ? [2 µF] Hint. is C. CAPACITANCE ∴ 4 3 4 πR ρ = N × π r 3 ρ 3 3 1/3 R=r×N or ∴ R = N1/3 r C = N1/3 or C = c × N1/3 c Now V = Q and v = q C c ∴ or V Q  c   1  =   ×   = (N ) ×  1/3  = N2/3 v C  C  N  V = v × N2/3 C C C C C C C B C 15. [9 µC] Hint.. One capacitor is between outer sphere and earth and the other capacitor is between outer sphere and inner earthed sphere. The branch AB then has 7 µF and A 5 µF 4 µF 6V Fig.1 × 0.46 × 10–10 F] Hint. between AB is 6 V.12 m 9 × 109 1 0. a = 0. The capacitance of first row. 5. C/8 .46 × 10–10 F 17. Find the capacitances in two possible arrangements Charge on inner sphere.1 9 21 × 10 −8 C 9 9 × 10 When the outer shell is removed.1 9 × 109 2.5 cm. a = 0.D. [On A = 200/3 µC. Initial charge on A.1 m. q = CV = × 100 = [M. the effective capacitance of branch AB is C AB = 7 × 3 21 = µF 7 + 3 10 21 63 ×6= µC 10 5 17 Total charge in branch AB.N. then charge on the inner sphere remains the same. across parallel combination = 6 − Charge on 5 µF capacitor = (5 × 10–6) × 18. The capacitance of air-filled spherical capacitor is ab C = 4πε0 b−a Here ∴ 4πε0 = C= 1 9 × 109 1 9 × 10 9 = C2 N–1 m–2.1 × 0. A capacitor is filled with two dielectrics of the same dimensions but of dielectric constants K1 and K2 respectively.CAPACITANCE 3 µF is series. q2 = C2V = (5 × 10–6) × 100 = 500 µC When the oppositely charged plates of A and B are connected together. q = CABV = P. The radii of the two spherical shells which form an air filled spherical capacitor are 10 cm and 10. the capacitance will change and is given by 1 1 × 0.].R. However. Two parallel plate capacitors A and B having capacitance of 1 µF and 5 µF are charged separately to the same potential of 100 V.D.105 − 0. across 3 µF capacitor = q 63 1 21 = × = volts 3 5 3 5 21 9 = volts 5 5 9 = 9 × 10–6 C = 9 µC 5 ∴ P.105 m × 0.105 2. What is the final potential of the charged inner shell ? [2100 V] Hint.1 = × 10 −8 F C ′ = 4πε 0 a = 9 9 9 × 10 ∴ Final potential of the inner charged shell is 21 9 q = × 10 −8 × −10 = 2100 V V′ = 9 C′ 10 20. q1 = C1V = (1 × 10–6) × 100 = 100 µC Initial charge on B. the outer shell is taken apart and removed. After the inner shell has been charged to a potential of 100 V. Now positive plate of A is connected to the negative plate of B and the negative plate of A is connected to the positive of B.1 = F 0. b = 0. On B = 1000/3 µC] Hint. Therefore. (i ) 2ε 0 A d K1 K 2 K1 + K 2 ( ii ) ε0 A ( K1 + K 2 ) 2d . the net charge is q = q2 – q1 = 500 – 100 = 400 µC Final potential difference = Net charge 400 × 10−6 200 = = V Net capacitance (1 + 5)10−6 3 200 200 200 = (1 × 10–6) × = µC 3 3 3 Final charge on A = C1 × Final charge on B = C2 × 200 200 1000 = (5 × 10–6) × = µC 3 3 3 19. Find the final charge on each capacitor. each with plate area A/2 and plate separation d i. ∴ C′ = C1 = C2 = K1 ε0 ( A / 2) K1 ε0 A = d 2d K 2 ε0 ( A / 2) K 2 ε0 A = d 2d The equivalent capacitance C ″ is given by.74 (i) is equivalent to two capacitors in series. (i) The arrangement shown in Fig. 5. C1 = C2 = K1 ε 0 A 2 K1 ε0 A = d /2 d K 2 ε0 A 2 K 2 ε 0 A = d /2 d The equivalent capacitance C′ is given by. 5. 1 1 1 d d d  1 1  = + = + = +   C ′ C1 C2 2 K1 ε0 A 2 K 2 ε0 A 2 ε0 A  K1 K 2  = d  K1 + K 2    2 ε0 A  K1 K 2  2 ε0 A  K1 K 2    d  K1 + K 2  (ii)The arrangement shown in Fig..74 (ii) is equivalent to two capacitors inparallel.74.. The two possible arrangements are shown in Fig.e.74 ()i i B K2 A/2 A A/2 CAPACITANCE d Hint. 5. C ′′ = C1 + C2 = K1 ε0 A K 2 ε0 A ε0 A + = ( K1 + K 2 ) 2d 2d 2d ∴ C ′′ = ε0 A ( K1 + K 2 ) 2d .18 A d/2 d/2 K1 d K2 K1 B () i Fig.each with plate area A and plate separation d/2 i. 5.e. 5. 166. [600V. Calculate the radius of a spherical conductor of capacitance 1 farad. In the circuit shown in Fig. [(i) 885 µF. Determine the maximum capacitance of this variable capacitor.2 cm radius and 1. potential difference of 900V is applied to this series combination. Find charge and p. In the circuit shown in Fig. If charge on capacitor plates is 0. each of 1µF.23. 3µF and 6µF respectively are connected in series across 500V d. Find p.d. 83. A = length × width] d 6. a plate area of 0. C = 0 . If a p. (ii) 0.[ 5µF] µ µ 15.7V.d. A parallel plate capacitor has plates 0.5 × 10–5 C/m2 (ii) 28 V (iii) 186667 V/m] 9. 1. In the circuit shown in Fig.25 B Fig. 300V] 10. 5. Two capacitors of capacitance 2 µF and 4 µF respectively are connected in series.d. between plates. The area of each plate is 0. How will you combine four capacitors. Three capacitors of 2µF. A d. What must be the plate area if the capacitance of the capacitor is to be 2F? [1130 km2] 7.3 mm separation. what charge will appear on the plates? [140 pF. find the equivalent capacitance between points A and B. Here d = 0. Calculate the capacitance between points A and B in Fig. 5. [500 µC. [(i) 0. 5. find the equivalent capacitance between points A and B. Can a metal sphere of radius 1 cm hold a charge of 1C ? [9 × 106 km] [No] 3. 5.CAPACITANCE 1 PROBLEMS FOR PRACTICE 1.5 and its width and thickness are 50 mm and 0. 5.23 Fig. [1µF] 2 µF 3 µF A 1 µF 1 µF 2 µF 4 µF 4 µF B Fig.001m.3V] 12. 17 nC] 8.24 16. What is the capacitance of the capacitor. of 120 V is applied across the plates. How can three capacitors of capacitance 3µF. The plates of a parallel plate capacitor are separated by a distance of 0. [3µF] µ 13. 6µF and 9µF respectively be connected to have a capacitance of 11µF? [3µF and 6µF in series with 9µF in parallel with both] µ µ µ 11.0885 µC] 5. supply.5 cm.5 µC.0015 m2 and separation between opposite plates is 0. [292 pF] . 5.1m2 and a dielectric of relative permittivity 3. 250V.c.15 mm apart. [4µF] µ 2 µF 4 µF 4 µF 5 µF A 2 µF 6 µF B 6 µF A 4 µF 5 µF 4 µF B 14. ε KA [Hint.75µF? [Three in parallel and fourth in series with this parallel combination] 17. Calculate the capacitance of a conducting sphere of radius 10 cm situated in air. A variable air capacitor has 11 movable plates and 12 stationary plates.26. How much charge is required to raise it to a potential of 1000 V ? [11 pF.05 mm respectively. Now area. across each capacitor. on each capacitor. Find the length of the paper used in a parallel plate capacitor of capacitance 2µF if the dielectric constant of the paper is 2.05 mm. (ii) p. A parallel plate capacitor has circular plates of 8.24.d.25. Calculate (i) capacitance of the capacitor and (ii) charge on each plate.c.5 mm apart and each of effective area 500 cm2 is connected to a 100 V battery. 5. and (iii) electric field intensity.1 × 10–8 C] 4. to have a net capacitance of 0.26 3 µF 2 µF 3 µF 3 µF 3 µF 3 µF 3 µF A 8 µF Fig. A capacitor consisting of two parallel plates 0. 2. find (i) electric flux density. find the equivalent capacitance between points A and B. Now a dielectric slab (K = 4) of thickness 2mm is placed between the plates. 8. Write a short note on polar and non-polar dielectrics. Derive an expression for the capacitance of an isolated spherical conductor. 2. The capacitors are joined through a conducting wire. 11. are connected in series. 12. [(i) 10–8 C (ii) 125 V (iii) 6. 3. What do you mean by dielectric strength of a dielectric? . [29. 16.25 × 10–7 J ] 23. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor.33 µF capacitance if area of each sheet of tin foil is 82 cm2. Derive an expression for the capacitance of a spherical capacitor. Show that energy stored by a capacitor is q2/2C where q is the final charge on the capacitor and C is its capacitance. 21. What do you mean by energy density of electric field? 14. [6 × 10–6 J] 20.2 mm thick and have relative permittivity 5. Show that half the energy supplied by the battery is lost as heat while charging the capacitor. 10. the mica sheets are 0.2 CAPACITANCE 18. The space between the plates of the capacitors is filled with a dielectric material of dielectric constant of 4.5 pF] 24. 6. What are the individual capaci2 1 µF . A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness 3d/4 where d is the separation of the plates. Write a short note on multiplate capacitor. Two parallel-plate capacitors. Define the SI unit of capacitance. 6 mm thick is introduced between the parallel plates of a capacitor of plate area 2 × 10–2 m2 and plate separation 0. How is the capacitance changed when the slab is inserted between the plates ? 4K C = C0 K +3 26. Derive an expression for the capacitance of a cylindrical capacitor. 13. It is charged to 200V and then the charging battery is removed. What do you mean by dielectric constant of a dielectric? 9. What is a parallel plate capacitor? Derive an expression for its capacitance. Find the capacitance. µF tances ? 3 3 22. each of capacitance 40 µF. 182 sheets of mica] 19. 5.0 mm holds a charge of 0. A parallel-plate capacitor having plate area 100 cm2 and separation 1. What is a capacitor? How does a capacitor store charge? 4. Obtain an expression for their equivalent capacitance. [11. the stored energy is 4J for the same potential difference. Write a short note on conductors and insulators. Three capacitors are connected in (i) series (ii) parallel. An ebonite plate (K = 3). What is the value of common potential? 15. Find the dielectric constant of the material filling the gap. Discuss the behaviour of metallic conductors in electric field. 7. An uncharged capacitor is connected to a battery. Two capacitors are connected in parallel and the energy stored is 18J when a potential difference of 6000 V is applied across the combination. 17. [32 µF] LONG/SHORT ANSWER QUESTIONS 1. Calculate (i) final charge on each plate (ii) final potential difference between plates (iii) final energy in the capacitor. A 600 pF capacitor is charged by a 200Vsupply.12 µC when connected to a 120 V battery. The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is 4mm. Two capacitors of capacitances C1 and C2 are charged to potentials V1 and V2 respectively. Calculate the number of sheets of tin foil and mica for a capacitor of 0.3] 25. Find the equivalent capacitance of the system.01m. Discuss the behaviour of a dielectric in a uniform electric field. How much electrostatic energy is lost in the process?. [183 sheets of tin foil. When the same capacitors are connected in series. Describe the principle. Why are capacitors connected in series or parallel? 9. 23. Show that SI unit of ε0 is F/m. 19. 5. Why is an insulator called a dielectric? 2. 11. Write a short note on lightning conductor. Write a short note on atmospheric electricity. Show that farad is a too large unit of capacitance. 12. 20. assuming that thickness of slab is less than the plate separation. How will you discharge a charged capacitor? 14. What is the difference between polar and non-polar dielectrics? 15. derive an expression for the capacitance of the resulting capacitor. What is the need of a multiplate capacitor? 10. If air is replaced by mica (K = 6). what is the capacitance of the capacitor? 16. Derive an expression for the capacitance of a parallel plate capacitor with conducting slab between the plates. Give an account of thunderstorms and lightning. Mention three advantages of dielectrics in capacitors. 6. 22. What is the dielectric constant of a metal? 17. Name the practical example of a cylindrical capacitor. construction and working of Van de Graaff generator. 21. How does lightning conductor prevent the building from lightning? 19. 24. A parallel plate air capacitor has plate separation d. What is the importance of energy density of electric field? 13. 4. Write a short note on the discharging effect of sharp points. A parallel plate air capacitor has a capacitance of 10 µF. How will you prove that electric field inside a charged conductor is zero? 3. If a dielectric slab of thickness t (t < d) is introduced between the plates. Why does charge leak off rapidly from pointed ends of a charged conductor? 20.CAPACITANCE 3 18. What is the importance of capacitance? 18. Show that electrostatic energy of a capacitor is stored in its electric field. Can you apply any value of electric field to a dielectric? 7. What is the breakdown voltage of air? . What do you mean by voltage rating of a capacitor? 8. VERY SHORT ANSWER QUESTIONS 1.

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