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CAPACITANCE 1
CONCEPTUAL QUESTIONS
Q. 1. The two charged conductors are touched mutually and then separated. What will be thecharge 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 areseparated further, what will be the effect on the reading of the voltmeter?
Ans. qVC
= and 0 ACd
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 conduct-
ing body connected to earth is also at zero potential or “ground potential”. Alternatively, thecapacitance of earth is so large that removal of electrons from it or supply of electrons to itmakes 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 isformed for free electrons which move across to discharge the surfaces. Stored electric poten-tial 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 thereany other advantage of solid dielectric?
Ans. There are other two advantages of a solid dielectric. First, it helps in keeping the plates closetogether 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 ACd
ε= or
38 2
120
(1 10 )(1) 1.1 10 m8.854 10
C dA−
−×= = × = ×
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? Bothspheres are of same radius.
Ans. Both the spheres will hold the same charge. It is because charge remains on the outer surface of acharged 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 willgive more intense electric shock if touched?
Ans. q = CV. Since V is constant, q ∝ C. It means that capacitor having large capacitance will store morecharge. Hence, when 1µF capacitor is touched, the discharging current will be high and you willget 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 bya 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, therewill be redistribution of charge (i.e., flow of charge through wire) till the two spheres attain thesame potential. Due to the flow of charge through the connecting wire, some energy will be lost asheat. 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 posi-tive charge?
Ans. Yes. We know that V = q/C. The capacitance depends upon the dimensions of the conductor. Ifthe two conductors are of different shapes and sizes, they will be charged to different potentialswhen 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 freeelectrons.
(ii) When a conductor is placed in an external electric field, there is no electric field inside theconductor. 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 withoutbreaking down.
Q. 12. Show that capacitance of a metal plate A can be increased by bringing anothermetal plate B near A.
Ans. The charge holding property of a conductor is called its capacitance. Fig. 5.47shows an insulated metal plate A. Let positive charge be given to it till its poten-tial becomes maximum. No further charge can be given to the plate as it wouldleak out. Now bring another insulated metal plate B near the plate A as shown inFig. 5.47. The plate A will induce negative charge on the inner face of B and equalpositive charge on the outer face. The induced positive charge tends to increasethe potential of plate A while induced negative charge tends to decrease the poten-tial of plate A. Since the induced negative charge is nearer to plate A than theinduced 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 maximumvalue. Thus the capacitance of conductor A is increased by bringing anotheruncharged conductor B near it.
Q. 13. In the above question, show that capacitance of metal plate A can be furtherincreased by earthing the plate B.
Ans. If the plate B is earthed [See Fig. 5.48], the induced positive charge being free willflow to earth. However, induced negative charge remains since it is bound to thepositive charge on plate A. As a result, the potential of plate A is sufficientlyreduced. We can now give a large amount of charge to plate A to raise itspotential to maximum value. We arrive at a very important conclusion that capacitance of aninsulated 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 nowbrought close to form a parallel-plate capacitor with capacitance C, what will be the poten-tial difference between the plates ?
Ans. Suppose A is the area of each plate. When the plates are held at a distance d, the capacitance ofthe parallel plate capacitor formed is
0 ACd
ε=
If E1 and E2 are the electric fields due to two plates, then net electric field E between the twoplates is
1 21 2 1 2
0 0 0
/ / 1 ( )2 2 2
q A q AE E E q qA
= − = − = −ε ε ε
∴ P.D. between plates, 1 2 1 20 0
1 ( ) ( )2 2
dV E d q q d q qA A
= = − × = −ε ε
1 2
2q q
C−= 0 AC
dε =
∵
Q. 15. A man fixes outside his house one evening a two metre high insulating slab carrying on its topa large aluminium sheet of area 1 m2. Will he get electric shock if he touches the metal sheetnext morning ?
Ans. Yes, the man gets shock. The discharging current in the atmosphere will charge the aluminiumsheet. 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 apotential difference of V. If after charging P, the battery is disconnected from it and the chargedcapacitor 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 = 2CTotal charge = q + 0 = q
∴ Potential difference V′ across each capacitor is
2 2 2q CV VVC 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 fromcapacitor 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 themetallic body of the car.
Q.18. Five identical capacitor plates each of area A are arrangedsuch that the adjacent plates are at a distance d apart. The platesare connected to a battery of V volts as shown in Fig. 5.49. Whatis 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 adja-cent plates is d, all the four capacitors have the same capacitance C. The plate 1 acts as apositive plate of capacitor C1 (of capacitance C ) formed between plates 1 and 2.
∴ Charge on plate 01 ACV Vd
ε= + =
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 024 ( ) AC C V Vd
− 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 capaci-tance of earth?
Solution: Capacitance of earth, C = 4π ε0 r
Here, 4πε0 91 ;
9 10=
× r = 6400 km = 6.4 × 106 m
∴6
96.4 109 10
C ×=×
= 0.711 × 10–3 F = 711 µµµµµF
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 104V?Solution:(i) Capacitance of sphere, C = 4 π ε0 r
∴ Radius of sphere, 9 12 2
0
1 (9 10 ) (50 10 ) 45 10 m4
r C − −= × = × × × = × =πε
45cm
(ii) Charge to be placed, q = CV = (50 × 10–12) × 104 = 5 × 10–7 C = 0.5 µµµµµCExample 3. Twenty seven spherical drops, each of radius 3mm and carrying 10–12C 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 34 4273 3
R rπ = × π
or R = 3r = 3 × 3 = 9 mm = 9 × 10–3 m
Capacitance of bigger drop, 3 120 9
14 9 10 10 F9 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
1227 10
10qVC
−
−×= = = 27 V
Example 4: A spherical capacitor has an inner sphere of radius 9 cm and an outer sphere of radius10cm. The outer sphere is earthed. Assume there is air in the space between the spheres. What is the capaci-tance of the capacitor?
Solution: Capacitance of the spherical capacitor is04
( )A B
B A
r rCr rπε
=−
Here rB = 10 cm = 0.1 m; rA = 9 cm = 0.09 m; 4π ε0 91
9 10=
×
∴9
0.09 0.19 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. Thecapacitor has the same capacitance as the capacitance of sphere of 1.2m diameter. Find the radii of itssurfaces.
Solution:
Given : 00
44A B
B A
r r Rr rπε
= πε−
A B
B A
r r Rr r
∴ =−
Here rB – rA = 2 cm and R = 1.2/2 = 0.6m = 60 cm
∴ 602
A Br r = or rA rB = 120 cm
Now (rB + rA)2 = (rB – rA)2 + 4 rA rB = (2)2 + 4 × 120 = 484
∴ 484 22cmB Ar r+ = =Since rB – rA = 2cm and rB + rA= 22cm, rB = 12cm ; rA= 10 cm
2 CAPACITANCE
Example 6. The plates of a parallel plate air capacitor are separated by a distance of 1 mm. What mustbe 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 ACd
ε=
Here d = 1 mm = 10–3 m; A = ?; C = 1F
∴3
120
1 108.854 10
CdA−
−×= =
ε ×= 1.1 × 108 m2
Note the enormous magnitude of plate area required to have a capacitance of 1F. This shows that faradis 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 aircapacitor 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 m2
Radius of sphere, r = 0.5 mFor parallel plate capacitor, C = ε0A/dFor spherical conductor, C = 4π ε0 r
Since the capacitance of the two capacitors is the same,
∴ 0 Ad
ε = 4πε0r
or 0.024 4 0.5
Adr
= =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; theplates being separated by a dielectric 2 mm thick and of relative permittivity 6. If the electric field strengthbetween the plates is 500 V/mm, calculate the charge on each plate.
Solution: Capacitance, 12
03
(8.854 10 )(6) (30)2 10
K ACd
−
−ε ×= =
× = 0.797 × 10–6 F = 0.797 µµµµµF
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 m2 separated by a dielectric 1 mm thick. The resulting capacitance of thearrangement is 300pF. Calculate (i) charge on each plate, (ii) electric flux density, (iii) potential gradi-ent, and (iv) relative permittivity of the dielectric.
Solution: C =300 pF = 300 × 10–12 F; V = 10 kV = 10 × 103 volts(i) Charge on each plate, q = CV = (300 × 10–12) (10 × 103) = 3 × 10–6 C
(ii) Electric flux density, 33 10
0.01qA
×σ = = = –4 23 × 10 C/m
(iii) Potential gradient, 3
310 101 10
VEd −
×= = =×
710 V/m
(iv) Electric intensity, 0
EK
σ=ε
∴ Relative permittivity, 4
12 70
3 108.854 10 10
KE
−
−σ ×= = =
ε × ×3.39
Example 10. A parallel plate capacitor is to be designed with a voltage rating 1kV using a materialof dielectric constant 3 and dielectric strength of 107 Vm–1. What minimum area of the plate is requiredto have a capacitance of 50 pF?
Solution. For reasons of safety, the electric field E between the plates should not exceed 10% of thedielectric strength of the dielectric i.e. E = 10% of 107 = 106 Vm–1.
Now VEd
= ∴3
36
1 10 10 m10
Vdd
−×= = =
∴ Plate area 12 3
12(50 10 ) 10
8.85 10 3A
− −
−× ×= =
× ×-3 21.9 × 10 m
CAPACITANCE 3
Example 11. Two parallel plate air capacitors have their plate areas 100 cm2 and 500 cm2 respec-tively. If they have the same charge and potential and the distance between the plates of the first capacitor is0.5 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 twocapacitors have the same charge and potential, their capacitances (C = q/V) are equal i.e. C1 = C2.
∴ 0 1 0 2
1 2
A Ad d
ε ε= ∴ 2
2 11
Ad dA
=
Here A2 = 500 cm2 ; A1 = 100 cm2 ; d1 = 0.5 mm = 0.05 cm
∴ 2500 0.05100
d = × = 0.25cm
Example 12. Three capacitors have capacitances of 0.5 µF, 0.3 µF and 0.2 µF respectively. They arefirst connected to have maximum capacitance and then connected to have minimum capacitance. Find theratio of maximum capacitance to minimum capacitance.
Solution: (i) For maximum capacitance, all the capacitors will have to be connected in parallel.CP = 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 1 310.5 0.3 0.2 3SC
= + + =
or CS = 3/31 µF
∴ P
S
CC
= 313
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 F15 20S
C CCC C
×= = = µ+ +
In series connection, charge on each capacitor is the same.∴ Charge on each capacitor, q = CS V = (8.57 × 10–6) × 600 = 5.14 × 10–3 C
(ii) P.D. across 15 µF capacitor 3
61
5.14 1015 10
qC
−
−×= = =
×342.7 V
P.D. across 20 µF capacitor 3
62
5.14 1020 10
qC
−
−×= = =
×257 V
Example 14. The total capacitance of two capacitors is 4µF when connected in series and 18 µFwhen connected in parallel. Find the capacitance of each capacitor.
Solution: Let C1 and C2 be the unknown capacitances. Then,C1 + C2 =18 ...(i) when in parallel
1 2
1 24C C
C C=
+ ...(ii) when in series
Multiplying eqs. (i) and (ii), C1C2 = 72
Now 21 2 1 2 1 2( ) 4C C C C C C− = + + 2(18) 4 72 6= − × = ± ... (iii)
Solving eqs. (i) and (iii), we get, C1 = 12 µµµµµF or 6 µµµµµF; C2 = 6 µµµµµF or 12 µµµµµFExample 15. In the circuit shown in Fig. 5.13, the total charge is 750 µC. Find the values of V1, V
and C2.
Solution: Voltage across6
1 1 61
(750 10 ),15 10
qC VC
−
−×= = =
×50V
Applied voltage, V = V1 + V2 = 50 + 20 = 70 VCharge on C3 = C3V2 = (8 × 10–6) 20 = 160 × 10–6 = 160 µC
∴ Charge on C2 = 750 – 160 = 590 µC
4 CAPACITANCE
Fig. 5.13
V+ _
V1
C2
C = 1 15 Fµ
V = 2 20 V
C = 3 8 Fµ
∴ Capacitance of 6
2590 10
20C
−×= = 29.5 × 10–6 F = 29.5 µµµµµF
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
C1
C2
C4
C3
200 pF 300 V
200 pF
100 pF
100 pF
+_
Fig. 5.15
300 V
+_
C3C2
C4
C1
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 equiva-lent capacitance C′ of series-connected capacitors C2 and C3 is
2 3
2 3
200 200 100pF200 200
C CCC C
× ×′ = = =+ +
The equivalent capacitance of parallel combination of C′ (= 100 pF) and C1 isCBC = C′ + C1 = 100 + 100 = 200 pF
The entire circuit now reduces to two capacitors C4 and CBC (= 200 pF) in series.∴ Equivalent capacitance of the network is
4
4
100 200100 200
BC
BC
C CCC C
× ×= = =+ +
200 pF3
Charges and p.d. on various Capacitors
Total charge, q = CV = 12 8200 10 300 2 10 C
3− − × × = ×
∴ Charge on C4 = 2 × 10–8 C
∴ P.D. across C4, V4 = 8
124
2 10100 10
qC
−
−×= =×
200V
P.D. between B and C, VBC = 300 – 200 = 100 VCharge on C1, q1 = C1 VBC = (100 × 10–12) × 100 = 10–8 C
P.D. across C1, V1 = VBC = 100 VP.D. across C2 = P.D. across C3 = 100/2 = 50 VCharge on C2 = Charge on C3 = Total charge – Charge on C1
= (2 × 10–8) – (10–8) = 10–8 C
CAPACITANCE 5
Example 17: Fig. 5.16 shows a network of four capacitors. Determine the equivalent capacitancebetween points A and B. If a 10V battery is connected between A and B, how much total charge will bestored on the capacitors.
Fig. 5.16C 3 = 9 Fµ
C 2 = 9 Fµ
C 1 = 3 Fµ
C 4 = 9 Fµ
A B
Fig. 5.17
C1
C2 C4C3
A B
Solution: The given network is equivalent to the network shown in Fig. 5.17. The equivalent capacitanceC′ of the series connected capacitors C2, C3 and C4 is given by;
2 3 4
1 1 1 1 1 1 1 19 9 9 3C C C C
= + + = + + =′
∴ C′ = 3 µFBetween points A and B, we now have capacitors C1 and C′ (= 3 µF) in parallel. Therefore, the equivalent
capacitance CAB between points A and B of the network isCAB = C1 + C′ = 3 + 3 = 6 µµµµµF
Total charge stored on the capacitors isq = CAB × V = (6 × 10–6) × 10 = 60 × 10–6 C = 60 µµµµµC
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 (plate2, plate 3 and plate 6) is connected to point B. Therefore,the three capacitors are in parallel. Hence, equivalentcapacitance between A and B is
CAB = C1 + C2 + C3Example 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) CAB = 10 µF
Fig. 5.19
8 F µ 8 F µ
12 F µ
10 F µ
400 V
8 F µ
8 F µ8 F µA B C D
+ _
CBC = 8 88 8 × +
+ (8) + (8) = 4 + 8 + 8 = 20 µF
10 20 20 F10 20 3
AB BCAC
AB BC
C CCC C
× ×= = = µ+ +
CCD = 8 + 12 = 20 µF
∴ (20 / 3) 20(20 / 3) 20ADC ×= = µ
+5 F
Fig. 5.18
C1 C2 C3
1 3 52 4 6A B
6 CAPACITANCE
(ii) Total charge, q = CAD × V = (5 × 10–6 ) × 400 = 0.002CThe total charge will divide between the parallel capacitors connected in the branch CD.
P.D. between C and D, 60.002 100 V
20 10CDCD
qVC −= = =
×∴ Charge on 12 µF capacitor = (12 × 10–6) × 100 = 1.2 × 10–3 C = 1.2 µµµµµCExample 20: In the network shown in Fig. 5.20 (i), C1 = C2 = C3 = C4 = 8 µF and C5 = 10 µF. Find
the equivalent capacitance between points A and B.
Fig. 5.20
C3 C3
C3
C5
C5
C4 C4
C4
C2 C2C2
C1
C1 C1
AD
CB
( )ii ( )iii( )i
AAB
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(C1C4 = C2C3 because C1 = C2 = C3 = C4), the bridge is balanced. It means that points C and D are at thesame potential. Therefore, there will be no charge on capacitor C5. Hence, this capacitor is ineffective andcan be removed from the circuit as shown in Fig. 5.20 (iii). Referring to Fig. 5.20 (iii), the equivalentcapacitance C′ of the series connected capacitors C1 and C2 is
1 2
1 2
8 8 4 F8 8
C CCC C
×′ = = = µ+ +
The equivalent capacitance C″ of series connected capacitors C3 and C4 [See Fig. 5.20 (iii)].
3 4
1 2
8 8 4 F8 8
C CCC C
×′′ = = = µ+ +
Now CAB = C′ || C″ = 4 || 4 = 4 + 4 = 8 µµµµµFExample 21: A mica dielectric parallel plate capacitor has 21 plates, each having an effective area
of 5 cm2 and each separated by a gap of 0.005mm. Find the capacitance of the capacitor. Take the relativepermittivity of mica as 6.
Solution: For a multiplate capacitor, capacitance is given by;
0( 1) K AC nd
ε= −
12 4
3(8.854 10 ) 6 (5 10 )(21 1)
0.005 10
− −
−× × × ×= −
×= 0.1062 × 10–6 F = 0.1062 µµµµµF
Example 22: A parallel plate capacitor has three similar parallel plates. Find the ratio of capaci-tance when the inner plate is mid-way between the outers to the capacitance when inner plate is threetimes 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. Thearrangement is equivalent to two capacitors in parallel. Capacitance of this capacitor,
0 0 01
4/ 2 / 2K A K A K AC
d d dε ε ε
= + =
Fig. 5.21( )ii( )i
d/2 3 /4d
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 C2 of this capacitor,
0 0 02
16/ 4 3 / 4 3K A K A K AC
d d dε ε ε
= + =
∴ C1/C2 = 0.75Example 23: Given some capacitors of 0.1µF capable of withstand-
ing 15V. Calculate the number of capacitors needed if it is desired toobtain 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, VC = 15 VSupply 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, CS = C/4 = 0.1/4 = 0.025 µF. Since it is desired to have atotal capacitance of 0.1 µF, number of such rows in parallel = C/CS = 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 betweenthe 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 0.210 F = 1041.4log ( / ) 41.4log (15.4 /15)
K lCb a
− −×= × ×
= 2.11 × 10–9F(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 CWhen the two capacitors are connected through a wire, the total capacitance CP = 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 60.0005
8 10P
qC −= =
× = 62.5 V
Example 26: Two capacitors of capacitance 4 µF and 6 µF respectively are connected in seriesacross a p.d. of 250V. The capacitors are disconnected from the supply and are reconnected in parallelwith 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 theircapacitances.
∴ P.D. across 4 µF capacitor 6250 150V4 6
= × =+
Charge on 4 µF capacitor = (4 × 10–6) × 150 = 0.0006 CSince the capacitors are connected in series, charge on each capacitor is the same.∴ Charge on both capacitors = 2 × 0.0006 = 0.0012 CParallel connection. When the capacitors are connected in parallel, the total capacitance CP = 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 6Total charge 0.0012
10 10PC −= =×
= 120 V
Charge on 4 µF capacitor = (4 × 10–6) × 120 = 480 × 10–6 = 480 µµµµµCCharge on 6 µF capacitor = (6 × 10–6) × 120 = 720 × 10–6 = 720 µµµµµCExample 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 C1 and C2 µF be the capacitances of the capacitors A and B respectively. When thecapacitors are connected in series [See Fig. 5.31 (i)], charge on each capacitor is the same.
Fig. 5.22
C CCC
+ _
60 V
8 CAPACITANCE
∴ C1 × 120 = C2 × 80or C2 = 1.5 C1 ...(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 (C2 + 3). Thus the circuit shown in Fig. 5.31 (ii) can be thought of as a series circuitconsisting of capacitances C1 and (C2 + 3) connected in series.
∴ C1 × 140 = (C2 + 3) 60or 7C1 – 3C2 = 9 ...(ii)
Fig. 5.31
140 V 60 V 120 V 80 V
C2C1
A B
+ + __
200 V 200 V ( )ii( )i
A B
C1 C2
3 F µ
Solving eqs. (i) and (ii), we have, C1 = 3.6 µµµµµF; C2 = 5.4 µµµµµFExample 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) theelectrostatic energies before and after the capacitors are connected.
Solution: C1 = 16 µF; C2 = 4 µFBefore joiningCharge on 16 µF capacitor, q = C1V1 = (16 × 10–6) × 100 = 1.6 × 10–3 C
Energy stored, 2 6 21 1 1
1 1 (16 10 ) 1002 2
U C V −= = × × = 0.08 J
After joiningWhen the capacitors are connected through a wire, the total capacitance, Cp = C1 + C2 = 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
61.6 1020 10P
qVC
−
−×= =×
= 80 V
Energy stored, 22
1 12 2PU 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 conductorconnecting the capacitors.
Example 29: A metal sphere 4 m in diameter is charged to a potential of 3 MV. Calculate the heatgenerated when the sphere is earthed through a long resistance wire.
Solution: Potential at the surface of sphere, 99 10 qVr
= ×
∴ Charge on sphere, 6
39 9
(3 10 ) 2 0.67 10 C9 10 9 10V rq −× × ×= = = ×× ×
Energy stored in sphere 3 61 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 = ε0 A/d) becomes
twice. Therefore, the charge on the capacitor (q′ = C′V)becomes twice. The extra charge is supplied bythe battery.
Extra charge supplied by battery = q′ – q = 2q – q = q = 1.35 × 10–10 CExample 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 thenew stored energy?
CAPACITANCE 9
Solution:C = 100µF = 100 × 10–6F = 10–4F; V = 500 voltsWhen 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. Thepotential difference between the plates must decrease to maintain the same charge.
∴ q = CV = C′V′ or 5002 2 2
CV CV VVC C
′ = = = = =′
250 volts
New stored energy 2
21 1 (2 )2 2 2
VC V C ′ ′= =
221 1 1
2 2 2 2CV CV = =
4 21 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 tubein 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 21 1 (2000 10 ) (1.5)2 2
U CV −= = × × ×
= 2.25 × 10–3 JIn order to produce the flash, the capacitor discharges
in 0.1 ms (= 0.1 × 10–3 s).
∴ Power of flash 3
32.25 10
Time 0.1 10U −
−×= = =
×22.5 W
Example 33: A parallel plate capacitor is partially filled with an ebonite plate of thickness 6 mm. Thearea of the plates of the capacitor is 2 × 10–3 m2 and the distance between them is 0.01m. The dielectricconstant for ebonite is 3. Calculate the capacitance of the capacitor.
Solution:
Capacitance of the capacitor, 01(1 )
ACd t
K
ε=
− −
Here A = 2 × 10–3 m2; d = 0.01 m; t = 6 × 10–3 m; K = 3
∴12 3
3(8.854 10 ) 2 100.01 6 10 (1 1/ 3)
C− −
−× × ×=
− × − = 2.95 × 10–12 F
Example 34: A parallel plate capacitor has plate area of 2 m2 spaced by three layers of differentdielectric materials. The relative permittivities are 2, 4, 6 and thicknesses are 0.5, 1.5 and 0.3 mm respec-tively. Calculate the capacitance of the capacitor.
Solution: Capacitance of capacitor, 0
31 2
1 2 3
AC tt tK K K
ε=
+ +
12
3 3 3(8.854 10 ) 2
0.5 10 1.5 10 0.3 102 4 6
−
− − −× ×=
× × ×+ + = 0.026 × 10–6 F
Example 35: A capacitor is composed of two plates separated by 3 mm dielectric of relative permit-tivity 4. An additional piece of insulation 5 mm thick is now inserted between the plates. If the capacitorhas now capacitance one-third of its original capacitance, find the relative permittivity of the additionaldielectric.
Solution: Fig. 5.40 (i) and Fig. 5.40 (ii) respectively show the two cases.
For the first case, 0 1 0 4K A ACd d
ε ε × ×= = ...(i)
For the second case, 0 03 3
1 2
1 2 2
3 3 10 5 104
A ACt tK K K
− −ε ε
= =× ×+ +
...(ii)
–+1.5 V 2000 Fµ Discharge
Tube
ElectronicTrigger
Fig. 5.32
10 CAPACITANCE
( )ii3 mm 5 mm
K 2 = ?K 1 = 4
( )i3 mm
K 1 = 4
Fig. 5.40
Dividing eq. (i) by eq. (ii), 2
4 3 533 4 K
= +
∴ K2 = 3.33Example 36: An air capacitor has two parallel plates of 1500 cm2 area and held 5 mm apart. If a
dielectric slab of area 1500 cm2, thickness 2 mm and relative permittivity 3 is now introduced between theplates, 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 capaci-tor, the capacitance is given by;
0
( / )AC
d t t Kε
=− − ...(i)
If the medium were totally air, the capacitance would have been
00
ACd
ε= ...(ii)
An inspection of eqs. (i) and (ii) reveals that with the introduction of the dielectric slab between theplates of air capacitor, its capacitance increases. The distance between the plates is effectively reduced byt – (t/K). In order to bring the capacitance to the original value, the plates must be separated by this muchdistance in air.
∴ New separation between plates = d + ( t – t /K) = 5 + ( 2 –2/3) = 6.33 mmExample 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) ofthickness 2 mm is introduced between the plates. Find (i) final charge on each plate, (ii) p.d. between theplates, (iii) final energy in the capacitor, and (iv) energy loss.
Solution: The capacitance of parallel plate air capacitor is given by;
00
ACd
ε= ...(i)
When a dielectric slab of thickness t is placed between the plates of the capacitor, capacitance is given by;
0m
AC td tK
ε=
− +...(ii)
∴ 0 ( / ) 4 2 (2 / 4) 54 8m
C d t t KC d
− + − += = =
Before Introduction of the Dielectric SlabWhen air capacitor is charged to 200 V and then battery is removed, charge (q) on each plate is
q = C0 × V0 = (50 × 10–12 ) × 200 = 10–8 CAfter Introduction of the Dielectric Slab
(i) Since the battery is removed before the introduction of the dielectric slab, the charge on capaci-tor 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 toCm and p.d. between plates decreases to V.
q = C0 V0 = Cm V
∴ 00
5 2008m
CV VC
= = × =
125V
CAPACITANCE 11
Note that p.d. between the plates decreases. This is in agreement with theory.(iii) Final energy stored in the capacitor,
81 1 (10 ) 1252 2
U qV −= = × = –76.25 × 10 J
(iv) Initial stored energy, U0 = 12
qV0 ; Final stored energy, U = 12
qV
∴ Energy loss = U0 – U = 12
q (V0 – V)
= 12
(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 atiny force on the slab and would do work on it equal to 3.75 × 10–7 J.
Note: Initial stored energy, U0 = 12
qV0 = 12
(10–8) × 200 = 10– 6 J
Final stored energy U = 6.25 × 10–7 JHad the slab been 4 mm thick (= distance between plates), the energy stored in the capacitor would have
been smaller by 1/K i.e. U0/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 micasheet of dielectric constant 6 were inserted between the plates (i) while the voltage supply remainsconnected (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 timesi.e.
C′ = KC = 6 × 17.7 = 106.2 pFThe 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. Chargeon capacitor, q′ = C V = 106.2 × 10–12 × 100 = 1.06 × 10–8 CThe extra charge is supplied by the battery.
(ii) After the voltage supply is disconnectedC′ = KC = 6 × 17.7 = 106.2 pF
Charge on capacitor, q = CV = 17.7 × 10–12 × 100 = 1.77 × 10–9 CSince the supply is disconnected, the charge on the plates remains the same. Because the capaci-tance (C = q/V) has increased, the potential difference across the plates must decrease to maintainthe same charge.
9
121.77 10
106.2 10qVC
−
−×′ = = =
′ ×16.67 V
Example 39. In a Van de Graaff generator, the shell electrode is at 25 × 105 V. The dielectric strengthof the gas surrounding the electrode is 5 × 107 V/m. Calculate the minimum radius of the spherical shell.
Solution. Electric potential V of a charged shell is
0
14
qVr
=πε
Electric field at the surface of a charged shell is
20
14
qEr
=πε
∴ V rE
= or5
27
25 10 5 10 m5 10
VrE
−×= = = × =×
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 µFcharged to 200 V. Find the common potential. [Roorkee]
[400 V]Hint. Charge on one capacitor, q1 = C1 V1 = (20 × 10–6) × 500 = 0.01 C
Charge on second capacitor, q2 = C2 V2 = (10 × 10–6) × 200 = 0.002 C Total charge on capacitors, q = q1 + q2 = 0.01 + 0.002 = 0.012 C Total capacitance, C = C1 + C2 = (20 × 10–6) + (10 × 10–6) = 30 × 10–6 F
∴ Common potential, 60.012 400 V
30 10qVC −= = =
×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,
4Cn
= or C = 4 n = 4 × 5 = 20 µF
When the capacitors are connected in parallel, then equivalent capacitance C′ isC′ = 5 × 20 = 100 µF = 100 × 10–6 F
Energy stored is given by;
12
U = C′ V2 = 12
× (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 ofthe paper is 2.5 and its width and thickness are 50 mm and 0.05 mm respectively. [90 m]
Hint. 0 K ACd
ε=
Here C = 2 µF = 2 × 10–6 F; K = 2.5; d = 0.05 mm = 0.05 × 10–3 m
∴6 3
212
0
(2 10 ) (0.05 10 ) 4.5m8.85 10 2.5
CdAK
− −
−× × ×= = =
ε × ×
∴ 3Area 4.5Length = m 90 m
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 µµµµµF]Hint. The common potential V after connection is
1 1 2 2
1 2
C V C VVC C
+=+
Here C1 = 5 µF; V = 3 volts; V1 = 12 volts, V2 = 0; C2 = ?
∴ 2
2
5 12 035
CC
× + ×=+
∴ C2 = 15µF5. 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 ofpaper is 16 × 166 V/m. [(i) 19.6 × 10–12 F (ii) 0.31 µµµµµC]
Hint. (i) 0 K ACd
ε=
Here ε0 = 8.85 × 10–12 C2 N–1 m–2; K = 3.7; A = 6 10–4 m2; d = 1 10–3 m12 4
123
(8.85 10 ) (3.7) (6 10 ) 19.6 10 F1 10
C− −
−−
× × × ×= = ××
(ii) Since the thickness of the paper is 1 mm, the maximum voltage that can be applied before break-down occurs is
CAPACITANCE 13
Vmax = Emax × dHere Emax = 16 × 106 V/m; d = 1 mm = 1 × 10–3 m∴ Maximum charge that can be placed on capacitor is
qmax = C Vmax = (19.6 × 10–12) × (16 × 103) = 0.31 × 10–6 C = 0.31 µC6. A capacitor of capacitance C1 = 1 µF withstands the maximum voltage V1 = 6 kV while
another capacitance C2 = 2 µF withstands the maximum voltage V2 = 4 kV. What maximumvoltage will the system of these two capacitors withstand if they are connected in series ?
[M.N.R. 1992] [9 kV]Hint. 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 CThe charge on capacitor C1 should not exceed 6 × 10–3 C. Therefore, when capacitors are con-nected in series, the maximum charge that can be placed on the capacitors is 6 × 10–3 C ( = q1).
∴3 3
1 16 6
1 2
6 10 6 101 10 2 10max
q qVC C
− −
− −× ×= + = +× ×
= 6 × 103 + 3 × 103 = 103 (6 + 3) = 9 × 103 V = 9 kV7. A parallel-plate capacitor is charged with a battery to a charge q0 as shown in Fig. 5.70 (i).
The battery is then removed and the space between the plates is filled with a dielectric ofdielectric constant K. Find the energy stored in the capacitor before and after the dielec-
tric is inserted.
2 20 0
0 0;
2 2q qC K C
Hint. Energy stored in the capacitor in the absence of dielectric is
20 0 0
12
U C V=
Since V0 = q0/C0, this can be expressed as :2
00
02q
UC
= ... (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 capacitorremains the same. But the capacitance of the capacitor is increased K times i.e., new capacitance is C′= K C0 [See Fig 5.70 (ii)].
Fig. 5.70
q0 q0
KC0
V0
+ +_ _
( )i ( )ii
C0
∴ Energy stored in the capacitor after insertion of dielectric is2 2
0 0 0
02 2q q U
UC K C K
= = =′
or 0UUK
= ... (ii)
Since K > 1, we find that final energy is less than the initial energy by the factor 1/K. How will youaccount for “missing energy” ? When the dielectric is inserted into the capacitor, it gets pulledinto the device. The external agent must do negative work to keep the dielectric from accelerating.This work is simply = U0 – U. Alternately, the positive work done by the system = U0 – U.
8. Suppose in the above problem, the capacitor is kept connected with the battery and thendielectric 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 V0 will remain unchanged. As a result,electric field ( = V0/d) will also remain unchanged.
14 CAPACITANCE
The capacitance C0 will increase to C = K C0The charge will also increase to q = K q0 as explained below. q0 = C0 V0 ; q = C V0 = KC0 V0 = K q0
Initial stored energy, 20 0 0
12
U C V=
Final stored energy, 2 20 0 0 0 0 0
1 12 2
U C V K C V KU= = =
∴ U = KU0Note that stored energy is increased K times. Will any work be done in inserting the dielectric? Theanswer is yes. In this case, the work will be done by the battery. The battery not only gives theincreased energy to the capacitor but also provides the necessary energy for inserting thedielectric.
9. A parallel plate capacitor is maintained at a certain potential difference. When a 3mm slab isintroduced between the plates, in order to maintain the same potential difference, the distancebetween 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 ACd
ε= ... (i)
With the introduction of slab of thickness t, the new capacitance is
0
(1 1/ )AC
d t Kε′ =
′ − − ... (ii)
Now the charge (q = CV) remains the same in the two cases.
∴ 0 0
(1 1/ )A A
d d t Kε 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 11K
−
or 2.4 × 10–3 = 3 × 10–3 11K
−
∴ K = 510. 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
C1 = 50 pF = 50 × 10–12 FWith dial at 180°, the capacitance of the capacitor is C2 = 950 pF = 950 × 10–12 F P.D. across C2, V2 = 400 V∴ Charge on C2, q = C2V2 = (950 × 10–12) × 400 = 380 × 10–9 CWhen the battery is disconnected, charge q remains the same. Suppose V1 is the potential differenceacross the capacitor when the dial reads 0°.∴ q = C1V1
or 9
1 121
380 10 7600V50 10
qVC
−
−×= = =
×(ii) Work required = Gain in energy of the capacitor
2 21 1 2 2
1 12 2
C V C V= −
12 2 12 21 150 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 aretransferred from one plate to the other ? [6.9 × 109]
Hint. q = CV = (90 × 10–12) × (12) = 1.1 × 10–9 CNow q = ne∴ Number of electrons transferred is
9
919
1.1 10 6.9 101.6 10
qne
−
−×= = = ××
12. If C1 = 20 µF, C2 = 30 µF and C3 = 15 µF and the insulated plate of C1 be at a potential of 90 V, oneplate of C3 potential of 90 V, one plate of C3 being earthed, what is the potential difference betweenthe plates of C2, three capacitors being connected in series ? [20 V]
Hint. Fig. 5.71 shows the conditions of the problem. The equivalent capacitance of this series combina-tion is given by;
1 2 3
1 1 1 1 1 1 1 320 30 15 20C C C C
= + + = + + =
620 20F 10 F3 3
C −= µ = ×
P.D. across series combination, V = 90 – 0 = 90 V
Total charge, q = CV = 203
× 10–6 ×90 = 6 × 10–4 C
∴ P.D. across 4
2 2 62
6 10, 20V30 10
qC VC
−
−×= = =×
13. A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. Theouter sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between theconcentric 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 thecapacitance 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;
04 a bC Kb a
= πε−
Here 0
14πε
= 9 × 109 C2N–1 m–2; K = 32; b = 0.13 m; a = 0.12 m
∴ 99
32 0.12 0.13 5.55 10 F0.13 0.129 10
C −×= × = ×−×
(ii) Potential of inner sphere is
6
29
2.5 10 4.5 10 V5.55 10
qVC
−
−×= = = ××
(iii) Capacitance of isolated sphere is
11
0 90.124 1.33 10 F
9 10r −= πε = = ×
×Note that capacitance of an isolated sphere is much smaller than that of the concentric spheres. It isbecause in case of concentric spheres, the total potential is distributed over two spheres and thepotential difference between the two spheres becomes smaller. Since capacitance (C = q/V) isinversely 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 N1/3 (iii) V = v N2/3]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 = NqThe capacitance of a spherical drop is proportional to the radius.
16 CAPACITANCE
∴ C Rc r
=
Since mass is a conserved quantity,
3 34 43 3
R N rπ ρ = × π ρ
R = r × N1/3
orRr
= N1/3
∴ Cc
= N1/3 or C = c × N1/3
Now andQ qV vC c
= =
∴ 1/31= ( ) ×V Q c N
v C C N = ×
= N2/3
or V = v × N2/3
15. An infinite identical capacitors each of capaci-tance 1 µF are connected as shown in Fig. 5.72.What is the equivalent capacitance between ter-minals A and B ? [2 µµµµµF]
Hint. It is clear from the figure that the rows of capaci-tors are connected in parallel. The capacitance offirst 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 8ABC C CC C C = + + + + = + + + +
1 2 2 1 2 F
1 1/ 2C 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
04 abC Kb a
= πε−
Here 0 914
9 10πε =
× = C2 N–1 m–2; K = 5; a = 0.1 m; b = 0.12 m
∴ 109
1 0.1 0.125 3.33 10 F0.12 0.19 10
C −×= × × = ×−×
(ii) When the inner sphere is earthed, the system is equivalent to capacitors in parallel. One capaci-tor is between outer sphere and earth and the other capacitor is between outer sphere and innerearthed sphere. The equivalent capacitance is
C = 4πε0 b + 4πε0 K ab
b a−
9 91 1 0.1 0.120.12
0.12 0.19 10 9 10×= × + ×
−× × = 0.13 × 10–10 + 3.33 × 10–10
= 3.46 × 10–10 F17. Find the charge on 5 µF capacitor in the circuit shown in
Fig. 5.73. [9 µµµµµC]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 equivalentcapacitance = 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
CC
A B
C C C
C
16 capacitors
Fig. 5.73
A B3 Fµ
2 Fµ
5 Fµ
4 Fµ
6 V
CAPACITANCE 17
3 µF is series. Therefore, the effective capacitance of branch AB is
7 3 21 F7 3 10ABC ×= = µ
+
Total charge in branch AB, q = CABV = 2110 × 6 =
635 µC
P.D. across 3 µF capacitor 63 1 21 volts3 5 3 5q= = × =
∴ P.D. across parallel combination 21 96 volts5 5
= − =
Charge on 5 µF capacitor = (5 × 10–6) × 95 = 9 × 10–6 C = 9 µC
18. Two parallel plate capacitors A and B having capacitance of 1 µF and 5 µF are chargedseparately to the same potential of 100 V. Now positive plate of A is connected to the negativeplate of B and the negative plate of A is connected to the positive of B. Find the final chargeon each capacitor. [On A = 200/3 µµµµµC; On B = 1000/3 µµµµµC]
Hint. Initial charge on A, q1 = C1V = (1 × 10–6) × 100 = 100 µC Initial charge on B, q2 = C2V = (5 × 10–6) × 100 = 500 µCWhen the oppositely charged plates of A and B are connected together, the net charge is q = q2 – q1 = 500 – 100 = 400 µC
6
6Net charge 400 10 200Final potentialdifference = V
Net capacitance 3(1 5)10
−
−×= =
+
Final charge on A = C1 × 200
3 = (1 × 10–6) ×
2003
= 2003
µC
Final charge on B = C2 × 200
3 = (5 × 10–6) ×
2003
= 1000
3 µC
19. The radii of the two spherical shells which form an air filled spherical capacitor are 10 cmand 10.5 cm. After the inner shell has been charged to a potential of 100 V, the outer shellis 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
04 abCb a
= πε−
Here 4πε0 = 91
9 10× = C2 N–1 m–2; a = 0.1 m; b = 0.105 m
∴ 9 9
1 0.1 0.105 2.1 F0.105 0.19 10 9 10
C ×= × =−× ×
Charge on inner sphere, 89
2.1 21100 10 C99 10
q CV −= = × = ××
When the outer shell is removed, then charge on the inner sphere remains the same. However, thecapacitance will change and is given by
8
0 91 14 0.1 10 F
99 10C a −′ = πε = × = ×
×∴ Final potential of the inner charged shell is
810
21 910 2100 V9 10
qVC
−−
′ = = × × =′
20. A capacitor is filled with two dielectrics of the same dimensions but of dielectric constants K1and K2 respectively. Find the capacitances in two possible arrangements
[M.N.R.].
0 01 21 2
1 2
2( ) ( ) ( + )
+ 2A AK Ki ii K K
d K K dε ε
18 CAPACITANCE
d
A
B
d/2
d/2
K1
K2
()i
d
A
B
K1 K2
()ii
A/2 A/2
Fig. 5.74Hint. 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 platearea A and plate separation d/2 i.e.,
1 0 1 01
2/ 2
K A K ACd d
ε ε= =
2 0 2 02
2/ 2
K A K ACd d
ε ε= =
The equivalent capacitance C′ is given by;
1 2 1 0 2 0 0 1 2
1 1 1 1 12 2 2
d d dC C C K A K A A K K
= + = + = + ′ ε ε ε
1 2
0 1 22K Kd
A K K += ε
∴ 0 1 2
1 2
2 A K KCd K K
ε′ = + (ii)The arrangement shown in Fig. 5.74 (ii) is equivalent to two capacitors inparallel,each with platearea A/2 and plate separation d i.e.,
1 0 1 01
( / 2)2
K A K ACd d
ε ε= =
2 0 2 02
( / 2)2
K A K ACd d
ε ε= =
The equivalent capacitance C ″ is given by;
1 0 2 0 01 2 1 2( )
2 2 2K A K A AC C C K K
d d dε ε ε′′ = + = + = +
∴ 01 2( )
2AC K K
dε′′ = +
CAPACITANCE 1
PROBLEMS FOR PRACTICE
1. Calculate the radius of a spherical conductor of capacitance 1 farad. [9 × 106 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 cm2 is
connected to a 100 V battery. Calculate (i) capacitance of the capacitor and (ii) charge on eachplate. [(i) 885 µµµµµF, (ii) 0.0885 µµµµµC]
5. Find the length of the paper used in a parallel plate capacitor of capacitance 2µF if the dielectricconstant of the paper is 2.5 and its width and thickness are 50 mm and 0.05 mm respectively.
[Hint. 0 .KACd
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 platearea if the capacitance of the capacitor is to be 2F? [1130 km2]
7. A parallel plate capacitor has circular plates of 8.2 cm radius and 1.3 mm separation. What is thecapacitance of the capacitor. If a p.d. of 120 V is applied across the plates, what charge will appearon the plates? [140 pF; 17 nC]
8. A parallel plate capacitor has plates 0.15 mm apart, a plate area of 0.1m2 and a dielectric of relativepermittivity 3. If charge on capacitor plates is 0.5 µC, find (i) electric flux density, (ii) p.d. betweenplates, and (iii) electric field intensity. [(i) 0.5 × 10–5 C/m2 (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. potentialdifference 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? [3µµµµµF and 6µµµµµF in series with 9µµµµµF 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 µµµµµC; 250V, 166.7V, 83.3V]12. Calculate the capacitance between points A and B in Fig. 5.23. [3µµµµµF]13. In the circuit shown in Fig. 5.24, find the equivalent capacitance between points A and B. [4µµµµµF]
Fig. 5.24Fig. 5.23
2 F µ
2 F µ6 F µ
6 F µ5 F µ5 F µ
4 F µ4 F µ
4 F µ4 F µ
A B
A B
14. In the circuit shown in Fig. 5.25, find the equivalent capacitance between points A and B.[ 5µµµµµF]15. In the circuit shown in Fig. 5.26, find the equivalent capacitance between points A and B. [1µµµµµF]
Fig. 5.26Fig. 5.25
2 Fµ
4 F µ
3 F µ
2 F µ 3 F µ2 F µ
3 F µ3 F µ3 F µ
3 F µ3 F µ
1 F µ
4 F µ
1 F µ8 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 is0.0015 m2 and separation between opposite plates is 0.001m. Determine the maximum capacitanceof 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 ofeach sheet of tin foil is 82 cm2, 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 theprocess?. [6 × 10–6 J]
20. An uncharged capacitor is connected to a battery. Show that half the energy supplied by thebattery is lost as heat while charging the capacitor.
21. Two capacitors are connected in parallel and the energy stored is 18J when a potential differ-ence of 6000 V is applied across the combination. When the same capacitors are connected inseries, the stored energy is 4J for the same potential difference. What are the individual capaci-
tances ?
2 1F ; F3 3
µ µ
22. The capacitance of a parallel plate capacitor is 50 pF and the distance between the plates is4mm. 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 eachplate (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 m2 and plate separation 0.01m. Find the capacitance. [29.5 pF]24. A parallel-plate capacitor having plate area 100 cm2 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 fillingthe gap. [11.3]
25. A slab of material of dielectric constant K has the same area as the plates of a parallel-platecapacitor but has a thickness 3d/4 where d is the separation of the plates. How is the capacitancechanged when the slab is inserted between the plates ?
26. Two parallel-plate capacitors, each of capacitance 40 µF, are connected in series. The spacebetween the plates of the capacitors is filled with a dielectric material of dielectric constant of4. Find the equivalent capacitance of the system. [32 µµµµµF]
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 equivalentcapacitance.
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 q2/2C 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 C1 and C2 are charged to potentials V1 and V2 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?
04=
+ 3KC C
K
CAPACITANCE 3
18. Derive an expression for the capacitance of a parallel plate capacitor with conducting slab betweenthe 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) isintroduced between the plates, derive an expression for the capacitance of the resultingcapacitor.
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 ε0 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 isthe 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?
