Chapter 7 Electrostatic Potential and Capacitance (Electrostatics Part 7) – Physics free study material by TEACHING CARE online tuition and coaching classes
Example: 117 Three capacitors of 2m f, 3m f and 6m f are joined in series and the combination is charged by means of a 24
volt battery. The potential difference between the plates of the 6m f capacitor is
[MP PMT 2002 Similar to MP PMT 1996]
(a) 4 volts (b) 6 volts (c) 8 volts (d) 10 volts
Solution: (a) Equivalent capacitance of the network is
1 = 1 + 1 + 1
Ceq
Ceq = 1mF
2 3 6
2mF
Q
3mF 6mF
Charge supplied by battery Q = C_{eq}.V Þ 1 × 24 = 24 mC
24 + –
Hence potential difference across 6m F capacitor =
6
= 4 volt.
24V
Example: 118 Two capacitors each of 1 m f capacitance are connected in parallel and are then charged by 200 V D.C. supply. The total energy of their charges in joules is [MP PMT 2002]
(a) 0.01 (b) 0.02 (c) 0.04 (d) 0.06
Solution: (c) By using formula U = 1 Ceq V ^{2}
2
1mF
Here
\
Ceq = 2mF
U = 1 ´ 2 ´ 10 ^{6} ´ (200)^{2}
2
= 0.04 J
1mF
+ – 200V
Example: 119 Five capacitors are connected as shown in the figure. The equivalent capacitance between the point A and B is
[MP PMT 2002; SCRA 1996; Pantnagar 1987]
(a) 1 m f (b) 2 m f (c) 3 m f (d) 4 m f
Solution: (b)
A Parallel
Series A
2 = 1mF
2 Þ
B
A
1mf
1mf
ß
2mf
Parallel
1 + 1 = 2mF
1mf
Series
1mf
B
1 + 1 = 2mF
Ü
1mf
B
2 = 1mF
2
Hence equivalent capacitance between A and B is 2mF.
Example: 120 In the following network potential difference across capacitance of 4.5 mF is [RPET 2001; MP PET 1992]
(a) 8 V (b) 4 V (c) 2 V (d) 6 V
Parallel

Solution: (a) Equivalent capacitance Ceq = 9 ´ 4.5 = 3mF
9 4.5
Charge supplied by battery Q = C_{eq} ´ V = 3 × 12 = 36 mC
Hence potential difference across 4.5 mF = 36 = 8V.
4.5
 mF
3mF
6mF
+ – 12 V
3 + 6 = 9mF
Example: 121 A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having dielectric constants K_{1}, K_{2} and K_{3} as shown in fig. If a single dielectric material is to be used to have the same capacitance C in this capacitor, then its dielectric constant K is given by [IIT Screening 2000]
(a)
1 = 1 + 1
K K1 K 2
+ 1
2K 3
(b)
1 = 1
K K1 + K 2
+ 1
2K 3
(c)
K = K1 K 2
K1 + K 2
+ 2K 3
(d)
K = K1
 K 2
+ 2K 3
Solution: (b) The effective capacitance is given by 1 = d é 1 + 1 ù


Ceq e _{0} A ê 2K 3 (K1 + K 2 )ú
The capacitance of capacitor with single dielectric of dielectric constant K is
C = Ke _{0} A
d
According to question Ceq
= C i.e.,
e _{0} A




é 1 1
= Ke _{0} A
ù d
Þ 1 =
K
1
2K3
+ 1 .
K1 + K 2
dê 2K
 K + K ú
Example: 122 Two capacitors C_{1} = 2mF and C_{2} = 6mF in series, are connected in parallel to a third capacitor C_{3} = 4mF. This arrangement is then connected to a battery of e.m.f. = 2 V, as shown in the fig. How much energy is lost by the battery in charging the capacitors ? [MP PET 2001]
(a)
22 ´ 10 ^{6} J
(b)
11 ´ 10 ^{6} J
(c)
æ 32 ö´ 10 ^{6} J

3
(d)
æ 16 ö´ 10 ^{6} J

3
è ø è ø
Solution: (b) Equivalent capacitance
Ceq
= C1C2
C1 + C2
 C3
= 2 ´ 6 + 4 = 5.5mF
8
\ U = 1 Ceq .V ^{2} = 1 ´ 5.5 ´ (2)^{2} = 11 ´ 10 ^{6} J
2 2
Example: 123 In the circuit shown in the figure, each capacitor has a capacity of 3mF. The equivalent capacity between A
and B is [MP PMT 2000]
Solution: (d)
(a)
3 mF
4
(b) 3 mF (c) 6 mF (d) 5 mF
3mF
3mF
Þ
A B A
3mF
3mF Þ
B A B
Hence equivalent capacitance
Ceq = 3 ´ 6 + 3 = 5mF.

3 6
3mF
Parallel
3 + 3 = 6mF
3mF
6mF
Example: 124 Given a number of capacitors labelled as 8mF, 250 V. Find the minimum number of capacitors needed to get an arrangement equivalent to 16 mF, 1000 V [AIIMS 2000]
(a) 4 (b) 16 (c) 32 (d) 64
Solution: (c) Let C = 8 mF, C¢ = 16 mF and V = 250 volt, V¢ = 1000 V
Suppose m rows of given capacitors are connected in parallel which each row contains n capacitor then
Potential difference across each capacitors V = V ‘
n
and equivalent capacitance of network C‘ = mC .
n
On putting the values, we get n = 4 and m = 8. Hence total capacitors = m × n = 8 × 4 = 32.
C‘ æ V ‘ ö ^{2} 16 æ 1000 ö ^{2}
Short Trick : For such type of problem number of capacitors n = C ´ ç V ÷
. Here n =
8 ç 250 ÷
= 32
è ø è ø
Example: 125 Ten capacitors are joined in parallel and charged with a battery up to a potential V. They are then disconnected from battery and joined again in series then the potential of this combination will be [RPET 2000]
(a) V (b) 10 V (c) 5 V (d) 2 V
Solution: (b) By using the formula V ‘ = nV Þ V ‘ = 10V.
Example: 126 For the circuit shown, which of the following statements is true [IITJEE 1999]
 With S_{1} closed, V_{1} = 15 V, V_{2} = 20 V (b) With S_{3} closed, V_{1} = V_{2} = 25 V
(c) With S_{1} and S_{2} closed V_{1} = V_{2} = 0 (d) With S_{1} and S_{3} closed V_{1} = 30 V, V_{2} = 20 V
Solution: (d) When S_{3} is closed, due to attraction with opposite charge, no flow of charge takes place through S_{3}. Therefore, potential difference across capacitor plates remains unchanged or V_{1} = 30 V and V_{2} = 20 V.
Alternate Solution
Charges on the capacitors are –
q_{1} = (30)(2) = 60 pC ,
q_{2} = (20)(3) = 60 pC or
q_{1} = q_{2} = q (say)
The situation is similar as the two capacitors in series are
+ q – + q –
2PF 3PF
q = 60pC q = 60pC
Þ + –
first charged with a battery of emf 50 V and then disconnected.
When S_{3} is closed, V_{1} = 30 V and V_{2} = 20 V.
+ – 50V
V_{1} = 30V V_{2} = 20V
Example: 127 A finite ladder is constructed by connecting several sections
of 2m F,4m F capacitor combinations as shown in the figure. It is terminated by a capacitor of capacitance C.
What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A
and B becomes independent of the number of sections in between
(a) 4 m F
(b)
C2
2m F
(c)
C1 = 4m F
18 m F
(d)
6 m F

Solution: (a) By using formula
C = 2
; C = 2m F
We get
C = 4 m F.
Example: 128 Figure shows two capacitors connected in series and joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors. [MP PMT 1999]
 C1 > C2
 C1 = C2
 C1 < C2
 The information is insufficient to decide the relation between C_{1} and C_{2}
Solution: (c) According to graph we can say that potential difference across the capacitor C_{1} is more than that across C_{2} .
Since charge Q is same i.e., Q = C V
= C V
Þ C1
= V2
Þ C < C
(V > V ).
1 1 2 2
C2 V1
1 2 1 2
Example: 129 Two condensers of capacity C and 2C are connected in parallel and these are charged upto V volt. If the battery is removed and dielectric medium of constant K is put between the plates of first condenser, then the potential at each condenser is [RPET 1998; IITJEE 1988]
(a)
V (b)
K + 2
2 + K
3V
(c)
2V (d)
K + 2
3V
K + 2
Solution: (d) Initially C
Q
Equivalent capacitance of the system Ceq = 3C
Total charge Q = (3C)V
Finally
V Equivalent capacitance of the system
KC Ceq = KC + 2C
Hence common potential
V = Q
(KC + 2C)
= 3CV =
(K + 2)C
3V .
K + 2
Example: 130 Condenser A has a capacity of 15 mF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1 mF with air between the plates. Both are charged separately by a battery of 100V. after charging, both are connected in parallel without the battery and the dielectric material being removed. The common potential now is [MNR 1994]
(a) 400V (b) 800V (c) 1200V (d) 1600V
Solution: (b) Charge on capacitor A is given by Q1 = 15 ´ 10 ^{6} ´ 100 = 15 ´ 10 ^{4} C
Charge on capacitor B is given by Q2 = 1 ´ 10 ^{6} ´ 100 = 10 ^{4} C
Capacity of capacitor A after removing dielectric = 15 ´ 10 6 = 1mF
15
Now when both capacitors are connected in parallel their equivalent capacitance will be C_{eq} = 1 + 1 = 2mF
So common potential = (15 ´ 10 4 ) + (1 ´ 10 4 ) = 800V.
2 ´ 10 ^{6}
Example: 131 A capacitor of
20mF
is charged upto 500V is connected in parallel with another capacitor of 10mF which is
charged upto 200V. The common potential is [CBSE 2000; CPMT 1999; BHU 1997]
(a) 500V (b) 400V (c) 300V (d) 200V




Solution: (b) By using V = C1 V1 + C2 V2 ; C = 20 mF, V = 500 V, C = 10 mF and V
= 200 V
C1 V2
V = 20 ´ 500 + 10 ´ 200 = 400V.
20 + 10
Example: 132 In the circuit shown [DCE 1995]
 The charge on C_{2} is greater than that of C_{1} (b) The charge on C_{2} is smaller than that of C_{1}
(c) The potential drop across C_{1} is smaller than C_{2} (d) The potential drop across C_{1} is greater than C_{2}
Solution: (d) Given circuit can be redrawn as follows
Ceq = 4 ´ 8 = 8 mF
12 3
So Q = 8 ´ 6 = 16mC
3
C_{1} = 4mF C2 = 8mF
Hence potential difference V1 = 16 = 4volt and V2 = 16 = 2 volt
i.e. V_{1} > V_{2}
4 8
Example: 133 As shown in the figure two identical capacitors are connected to a battery of V volts in parallel. When
capacitors are fully charged, their stored energy is
U_{1} . If the key K is opened and a material of dielectric
constant K = 3 is inserted in each capacitor, their stored energy is now U _{2} .
U1 will be [IIT 1983]
U 2
(a)
3 (b)
5
5 (c) 3 (d) 1
3 3
Solution: (a) Initially potential difference across both the capacitor is same hence energy of the system is
U1 = 1 CV ^{2} + 1 CV ^{2} = CV ^{2}
……..(i)
2 2
In the second case when key K is opened and dielectric medium is filled between the plates, capacitance of both the capacitors becomes 3C, while potential difference across A is V and potential difference across B is
V hence energy of the system now is
3
1 _{2} 1 æ V ö ^{2} 10 _{2}
…….(ii)
U _{2} = 2 (3C)V + 2 (3C) ç 3 ÷
= 6 CV
So,
è ø
U_{1} = 3
U 2 5
Example: 134 In the following figure the resultant capacitance between A and B is 1mF . The capacitance C is [IIT 1977]
(a)
32 mF
11
(b)
11 mF
32
(c)
23 mF
32
(d)
32 mF
23
Solution: (d) Given network can be simplified as follows
C 1mF
A
C 1mF
A
Parallel
2 + 2 = 4mF
8mF 6mF
12mF
4mF Þ
Series
6 ´ 12
8mF
4mF
4mF
Parallel
2mF
C
2mF
Parallel
= 4 mF
6 + 12
B
4mF
Series
8 ´ 4 8
8 + 4
4 + 4 = 8mF
B
A 8 + 8 = 32 mF A
= 8 mF

3 9 9 9

8 mF 8 mF
3 9
ß B B
A B
C 32 mF
9
Given that equivalent capacitance between A and B i.e., C_{AB} = 1mF
But
C ´ 32
CAB = 9
C ´ 32
hence 9 = 1 Þ C = 32 mF.
C + 32
9
C + 32 23
9
Example: 135 A
1mF capacitor and a
2mF capacitor are connected in parallel across a 1200 volts line. The charged
capacitors are then disconnected from the line and from each other. These two capacitors are now connected to each other in parallel with terminals of unlike signs together. The charges on the capacitors will now be
(a)
1800mC each (b)
400mC and 800mC
(c)
800mC and 400mC (d) 800mC and 800mC
Solution: (b) Initially charge on capacitors can be calculated as follows
Q1 + – 1mF
Q_{1} = 1 × 1200 = 1200 mC and Q_{2} = 2 × 1200 = 2400 mC
C1 Finally when battery is disconnected and unlike plates are
Q2 + – 2mF
C2
connected together then common potential
2400 – 1200
V ‘ = Q2 – Q1
C1 + C2
+ – 1200 V
= 1 + 2
= 400V
Hence, New charge on C_{1} is 1 ´ 400 = 400mC
And New charge on C_{2} is 2 ´ 400 = 800mC.
Example: 136 The two condensers of capacitances
2mF and
3mF
are in series. The outer plate of the first condenser is at
1000 volts and the outer plate of the second condenser is earthed. The potential of the inner plate of each condenser is
(a) 300 volts (b) 500 volts (c) 600 volts (d) 400 volts
Solution: (d) Here, potential difference across the combination is VA – VB = 1000V
Equivalent capacitance C
= 2 ´ 3 = 6 mF
2mF 3mF 0 V
^{eq} 2 + 3 5
+ 1000 V
A C
Hence, charge on each capacitor will be Q = Ceq ´ (VA – VB )
= 6 ´ 1000 = 1200mC
5
So potential difference between A and C, VA – VC = 1200 = 600V
2
Þ 1000 – VC = 600 Þ
Vc = 400V
Example: 137 Four identical capacitors are connected in series with a 10V battery as shown in the figure. The point N is earthed. The potentials of points A and B are
(a)
10V,0V
(b)
7.5V – 2.5V
(c)
5V – 5V
(d)
7.5V,2.5V
Solution: (b) Potential difference across each capacitor will be 10 = 2.5V
4
Hence potential difference between A & N i.e.,
VA – VN
= 2.5 + 2.5 + 2.5 = 7.5V
Þ VA – 0 = VA = 7.5V
While
VN – VB = 2.5
Þ 0 – VB = 2.5
Þ VB = 2.5V
Example: 138 In the figure below, what is the potential difference between the points A and B and between B and C
respectively in steady state [IITJEE 1979]
(a) 100 volts both (b)
VAB = 75 volts, VBC = 25 volts
(c)
VAB = 25 volts, VBC = 75 volts (d)
VAB = 50 volts VBC = 50 volts
Solution: (c) In steady state No current flows in the given circuit hence resistances can be eliminated
Parallel 3+3= 6mF
Parallel 1+1= 2mF
 6mF
 2mF C
Line (1)
V1=VAB V2=VBC
Þ
1mF
Line (2)
100 V
A + – C
100 V
By using the formula to find potential difference in series combination of two capacitor
æ æ C2 ö
C1 ö
ç V1 = ç ÷.V and V2 = V ÷


ç è C1 + C2 ø
C2 + C2 ÷
V = V
= æ 2 ö ´ 100 = 25V ;
V = V
= æ 6 ö ´ 100 = 75V.
 AB
ç 2 + 6 ÷
 BC
ç 2 + 6 ÷
è ø è ø
Example: 139 A capacitor of capacitance 5mF is connected as shown in the figure. The internal resistance of the cell is 0.5W.
The amount of charge on the capacitor plate is [MP PET 1997]
(a)
0mC
(b)
5mC
(c)
10mC
(d)
25mC
Solution: (c) In steady state current drawn from the battery i =
2.5
(1 + 1 + 0.5)
= 1A
Line (1)
In steady state capacitor is fully charged hence No current will flow through line (2)
Line (2)
Hence potential difference across line (1) is V = 1 ´ 2 = 2volt , the
same potential difference appears across the capacitor, so charge on capacitor Q = 5 ´ 2 = 10mC
Example: 140 When the key K is pressed at time t = 0 . Which of the following statements about the current i in the resistor AB of the adjoining circuit is true [CBSE 1995]
 i = 2mA at all t (b) i oscillates between 1mA and 2mA
(c) i = 1mA at all t (d) At t = 0, i = 2mA and with time it goes to 1mA
Solution: (d) At
t = 0 whole current passes through capacitance; so effective resistance of circuit is
1000W
and current
i = 2
1000
= 2 ´ 10 ^{3} A = 2mA . After sufficient time, steady state is reached; then there is no current in
capacitor branch; so effective resistance of circuit is
1000 + 1000 = 2000W
and current
i = 2
2000
= 1 ´ 10 ^{–}^{3} A = 1mA i.e., current is 2mA at t = 0 and with time it goes to 1mA .
Example: 141 The plates of a capacitor are charged to a potential difference of 320 volts and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 second the potential difference between the plates of the capacitor is 240 volts, then after 2 and 3 seconds the potential difference between the plates will be [MP PET 1998]
(a) 200 and 180 volts (b) 180 and 135 volts (c) 160 and 80 volts (d) 140 and 20 volts
Solution: (b) During discharging potential difference across the capacitor falls exponentially as V = V_{0}e ^{–l} ^{t} (l = 1/RC)
Where V = Instantaneous P.D. and V_{0} = max. P.D. across capacitor
After 1 second V_{1} = 320 (e^{–}^{l}) Þ 240 = 320 (e^{–}^{l}) Þ e ^{–l} = 3
4
_{l} æ 3 ö ^{2}
After 2 seconds V_{2} = 320 (e^{–} )^{2} Þ 320 ´ ç ÷ = 180 volt
è 4 ø
_{l} æ 3 ö ^{3}
After 3 seconds V_{3} = 320 (e^{–} )^{3} = 320 ´ ç ÷ = 135 volt
è 4 ø
Example: 142 Five similar condenser plates, each of area A. are placed at equal distance d apart and are connected to a source of e.m.f E as shown in the following diagram. The charge on the plates 1 and 4 will be
(a)
e _{0} A , 2e _{0} A
(b)
e _{0} AV , 2e _{0} AV
(c)
e _{0} AV , 3e _{0} AV
(d)
e _{0} AV , 4e _{0} AV
d d d d d d d d
Solution: (b) Here five plates are given, even number of plates are connected together while odd number of plates are connected together so, four capacitors are formed and they are in parallel combination, hence redrawing the figure as shown below.
Capacitance of each
e A
Capacitor is C = ^{ } ^{0}
d
Potential difference across each capacitor is V
e _{0} A
So charge on each capacitor Q = d V
e AV
Charge on plate (1) is + ^{ } ^{0}
d
While charge on plate 4 is
– e 0 AV ´ 2
d
= – 2e _{0} AV . V
d
Example: 143 Four plates are arranged as shown in the diagram. If area of each plate is A and the distance between two neighbouring parallel plates is d, then the capacitance of this system between A and B will be
(a)
4e _{0} A
d
(b)
3e _{0} A
d
(c)
2e _{0} A
d
(d)
e _{0} A d
Solution: (c) To solve such type of problem following guidelines should be follows
A B
Guideline 1. Mark the number (1,2,3……… ) on the plates
1 2
A B
Guideline 2. Rearrange the diagram as shown below
Guideline 3. Since middle capacitor having plates 2, 3 is short circuited so it should be eliminated from the circuit
Hence equivalent capacitance between A and B CAB
1 2
A B
= 2 e _{0} A
d