Chapter 31 : Capacitors - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

JEE : Chapter 31 : Capacitors - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


31.1
CHAPTER – 31
CAPACITOR
1. Given that
Number of electron = 1 × 10
12
Net charge Q = 1 × 10
12
× 1.6 × 10
–19
= 1.6 × 10
–7
C
? The net potential difference = 10 L.
? Capacitance – C = 
v
q
= 
10
10 6 . 1
7 ?
?
= 1.6 × 10
–8
F.
2. A = ?r
2
= 25 ?cm
2
d = 0.1 cm
c = 
d
A
0
?
= 
1 . 0
14 . 3 25 10 854 . 8
12
? ? ?
?
= 6.95 × 10
–5
?F.   
3. Let the radius of the disc = R
?Area = ?R
2
C = 1 ?
D = 1 mm = 10
–3
m
? C = 
d
A
0
?
? 1 = 
3
2 12
10
r 10 85 . 8
?
?
? ? ?
? r
2
= 
? ?
?
?
85 . 8
10 10
12 3
= 
784 . 27
10
9
= 5998.5 m = 6 Km
4. A = 25 cm
2
  = 2.5 × 10
–3
cm
2
d = 1 mm = 0.01 m
V = 6V Q = ?
C = 
d
A
0
?
= 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
Q = CV = 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
× 6 = 1.32810 × 10
–10
C
W = Q × V = 1.32810 × 10
–10
× 6 = 8 × 10
–10
J. 
5. Plate area A = 25 cm
2
  = 2.5 × 10
–3
m
Separation d = 2 mm = 2 × 10
–3
m
Potential v = 12 v
(a) We know C = 
d
A
0
?
  =  
3
3 12
10 2
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 11.06 ×10
–12
F
C = 
v
q
? 11.06 ×10
–12
= 
12
q
? q
1
= 1.32 × 10
–10
C. 
(b) Then d = decreased to 1 mm 
? d = 1 mm = 1 × 10
–3 
m
C = 
d
A
0
?
  = 
v
q
= 
3
3 12
10 1
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 
12
2
? q
2
= 8.85 × 2.5 × 12 × 10
–12
= 2.65 × 10
–10
C.
? The extra charge given to plate = (2.65 – 1.32) × 10
–10
= 1.33 × 10
–10
C.
6. C
1
= 2 ?F, C
2
= 4 ?F , 
C
3
= 6 ?F V = 12 V
cq = C
1
+ C
2
+ C
3
= 2 + 4 + 6 = 12 ?F = 12 × 10
–6
F
q
1
= 12 × 2 = 24 ?C, q
2
= 12 × 4 = 48 ?C, q
3
= 12 × 6 = 72 ?C 
5 cm
0.1 cm
1 mm
C 1 V C 2
C 3
Page 2


31.1
CHAPTER – 31
CAPACITOR
1. Given that
Number of electron = 1 × 10
12
Net charge Q = 1 × 10
12
× 1.6 × 10
–19
= 1.6 × 10
–7
C
? The net potential difference = 10 L.
? Capacitance – C = 
v
q
= 
10
10 6 . 1
7 ?
?
= 1.6 × 10
–8
F.
2. A = ?r
2
= 25 ?cm
2
d = 0.1 cm
c = 
d
A
0
?
= 
1 . 0
14 . 3 25 10 854 . 8
12
? ? ?
?
= 6.95 × 10
–5
?F.   
3. Let the radius of the disc = R
?Area = ?R
2
C = 1 ?
D = 1 mm = 10
–3
m
? C = 
d
A
0
?
? 1 = 
3
2 12
10
r 10 85 . 8
?
?
? ? ?
? r
2
= 
? ?
?
?
85 . 8
10 10
12 3
= 
784 . 27
10
9
= 5998.5 m = 6 Km
4. A = 25 cm
2
  = 2.5 × 10
–3
cm
2
d = 1 mm = 0.01 m
V = 6V Q = ?
C = 
d
A
0
?
= 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
Q = CV = 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
× 6 = 1.32810 × 10
–10
C
W = Q × V = 1.32810 × 10
–10
× 6 = 8 × 10
–10
J. 
5. Plate area A = 25 cm
2
  = 2.5 × 10
–3
m
Separation d = 2 mm = 2 × 10
–3
m
Potential v = 12 v
(a) We know C = 
d
A
0
?
  =  
3
3 12
10 2
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 11.06 ×10
–12
F
C = 
v
q
? 11.06 ×10
–12
= 
12
q
? q
1
= 1.32 × 10
–10
C. 
(b) Then d = decreased to 1 mm 
? d = 1 mm = 1 × 10
–3 
m
C = 
d
A
0
?
  = 
v
q
= 
3
3 12
10 1
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 
12
2
? q
2
= 8.85 × 2.5 × 12 × 10
–12
= 2.65 × 10
–10
C.
? The extra charge given to plate = (2.65 – 1.32) × 10
–10
= 1.33 × 10
–10
C.
6. C
1
= 2 ?F, C
2
= 4 ?F , 
C
3
= 6 ?F V = 12 V
cq = C
1
+ C
2
+ C
3
= 2 + 4 + 6 = 12 ?F = 12 × 10
–6
F
q
1
= 12 × 2 = 24 ?C, q
2
= 12 × 4 = 48 ?C, q
3
= 12 × 6 = 72 ?C 
5 cm
0.1 cm
1 mm
C 1 V C 2
C 3
Capacitor
31.2
7.
? The equivalent capacity.
C = 
2 1 3 1 3 2
3 2 1
C C C C C C
C C C
? ?
= 
30 20 40 20 40 30
40 30 20
? ? ? ? ?
? ?
= 
2600
24000
= 9.23 ?F
(a) Let Equivalent charge at the capacitor = q
C = 
V
q
? q = C × V = 9.23 × 12 = 110 ?C on each.
As this is a series combination, the charge on each capacitor is same as the equivalent charge 
which is 110 ?C. ?
(b) Let the work done by the battery = W
? V = 
q
W
? W = Vq = 110 × 12 × 10
–6
= 1.33 × 10
–3
J.
8. C
1
= 8 ?F, C
2
= 4 ?F , C
3
= 4 ?F
Ceq = 
3 2 1
1 3 2
C C C
C ) C C (
? ?
? ?
= 
16
8 8 ?
= 4 ?F
Since B & C are parallel & are in series with A
So, q
1
= 8 × 6 = 48 ?C q
2
= 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 
9. (a)
? C
1
, C
1
are series & C
2
, C
2
are series as the V is same at p & q. So no current pass through p & q.
2 1
C
1
C
1
C
1
? ? ?
2 1
C C
1 1
C
1 ?
?
C
p
= 
2
C
1
=
2
4
= 2 ?F
And C
q
= 
2
C
2
= 
2
6
= 3 ?F
? C = C
p
+ C
q
= 2 + 3 = 5 ?F
(b) C
1
= 4 ?F, C
2
= 6 ?F,
In case of p & q, q = 0
? C
p 
= 
2
C
1
= 
2
4
= 2 ?F
C
q
= 
2
C
2
= 
2
6
= 3 ?F
& C ? = 2 + 3 = 5 ?F
C & C ? = 5 ?F
? The equation of capacitor  C = C ? + C ?? = 5 + 5 = 10 ?F
12 V
8 ?F ?
A
4 ?F ?
B C
4 ?F ?
A
B
C 1
C 2
C 1
C 2
C 1 = 4
C 2 = 6
A
p
B
C 1
C 1
C 1 C 1
q
R
S
C 2
C 2
C 2
C 2
30 ?F ?
V = 12 V
40 ?F ? 20 ?F ?
Page 3


31.1
CHAPTER – 31
CAPACITOR
1. Given that
Number of electron = 1 × 10
12
Net charge Q = 1 × 10
12
× 1.6 × 10
–19
= 1.6 × 10
–7
C
? The net potential difference = 10 L.
? Capacitance – C = 
v
q
= 
10
10 6 . 1
7 ?
?
= 1.6 × 10
–8
F.
2. A = ?r
2
= 25 ?cm
2
d = 0.1 cm
c = 
d
A
0
?
= 
1 . 0
14 . 3 25 10 854 . 8
12
? ? ?
?
= 6.95 × 10
–5
?F.   
3. Let the radius of the disc = R
?Area = ?R
2
C = 1 ?
D = 1 mm = 10
–3
m
? C = 
d
A
0
?
? 1 = 
3
2 12
10
r 10 85 . 8
?
?
? ? ?
? r
2
= 
? ?
?
?
85 . 8
10 10
12 3
= 
784 . 27
10
9
= 5998.5 m = 6 Km
4. A = 25 cm
2
  = 2.5 × 10
–3
cm
2
d = 1 mm = 0.01 m
V = 6V Q = ?
C = 
d
A
0
?
= 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
Q = CV = 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
× 6 = 1.32810 × 10
–10
C
W = Q × V = 1.32810 × 10
–10
× 6 = 8 × 10
–10
J. 
5. Plate area A = 25 cm
2
  = 2.5 × 10
–3
m
Separation d = 2 mm = 2 × 10
–3
m
Potential v = 12 v
(a) We know C = 
d
A
0
?
  =  
3
3 12
10 2
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 11.06 ×10
–12
F
C = 
v
q
? 11.06 ×10
–12
= 
12
q
? q
1
= 1.32 × 10
–10
C. 
(b) Then d = decreased to 1 mm 
? d = 1 mm = 1 × 10
–3 
m
C = 
d
A
0
?
  = 
v
q
= 
3
3 12
10 1
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 
12
2
? q
2
= 8.85 × 2.5 × 12 × 10
–12
= 2.65 × 10
–10
C.
? The extra charge given to plate = (2.65 – 1.32) × 10
–10
= 1.33 × 10
–10
C.
6. C
1
= 2 ?F, C
2
= 4 ?F , 
C
3
= 6 ?F V = 12 V
cq = C
1
+ C
2
+ C
3
= 2 + 4 + 6 = 12 ?F = 12 × 10
–6
F
q
1
= 12 × 2 = 24 ?C, q
2
= 12 × 4 = 48 ?C, q
3
= 12 × 6 = 72 ?C 
5 cm
0.1 cm
1 mm
C 1 V C 2
C 3
Capacitor
31.2
7.
? The equivalent capacity.
C = 
2 1 3 1 3 2
3 2 1
C C C C C C
C C C
? ?
= 
30 20 40 20 40 30
40 30 20
? ? ? ? ?
? ?
= 
2600
24000
= 9.23 ?F
(a) Let Equivalent charge at the capacitor = q
C = 
V
q
? q = C × V = 9.23 × 12 = 110 ?C on each.
As this is a series combination, the charge on each capacitor is same as the equivalent charge 
which is 110 ?C. ?
(b) Let the work done by the battery = W
? V = 
q
W
? W = Vq = 110 × 12 × 10
–6
= 1.33 × 10
–3
J.
8. C
1
= 8 ?F, C
2
= 4 ?F , C
3
= 4 ?F
Ceq = 
3 2 1
1 3 2
C C C
C ) C C (
? ?
? ?
= 
16
8 8 ?
= 4 ?F
Since B & C are parallel & are in series with A
So, q
1
= 8 × 6 = 48 ?C q
2
= 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 
9. (a)
? C
1
, C
1
are series & C
2
, C
2
are series as the V is same at p & q. So no current pass through p & q.
2 1
C
1
C
1
C
1
? ? ?
2 1
C C
1 1
C
1 ?
?
C
p
= 
2
C
1
=
2
4
= 2 ?F
And C
q
= 
2
C
2
= 
2
6
= 3 ?F
? C = C
p
+ C
q
= 2 + 3 = 5 ?F
(b) C
1
= 4 ?F, C
2
= 6 ?F,
In case of p & q, q = 0
? C
p 
= 
2
C
1
= 
2
4
= 2 ?F
C
q
= 
2
C
2
= 
2
6
= 3 ?F
& C ? = 2 + 3 = 5 ?F
C & C ? = 5 ?F
? The equation of capacitor  C = C ? + C ?? = 5 + 5 = 10 ?F
12 V
8 ?F ?
A
4 ?F ?
B C
4 ?F ?
A
B
C 1
C 2
C 1
C 2
C 1 = 4
C 2 = 6
A
p
B
C 1
C 1
C 1 C 1
q
R
S
C 2
C 2
C 2
C 2
30 ?F ?
V = 12 V
40 ?F ? 20 ?F ?
Capacitor
31.3
10. V = 10 v 
Ceq = C
1
+C
2
[ ? They are parallel]
= 5 + 6 = 11 ?F
q = CV = 11 × 10 110 ?C ?
11. The capacitance of the outer sphere = 2.2 ?F
C = 2.2 ?F
Potential, V = 10 v
Let the charge given to individual cylinder = q.
C = 
V
q
? q = CV = 2.2 × 10 = 22 ?F
? The total charge given to the inner cylinder = 22 + 22 = 44 ?F
12. C = 
V
q
, Now V = 
R
Kq
So, C
1
= 
? ?
1
R / Kq
q
= 
K
R
1
= 4 ??
0
R
1
Similarly c
2
= 4 ??
0
R
2
The combination is necessarily parallel.
Hence Ceq = 4 ??
0
R
1
+4 ??
0
R
2
= 4 ??
0
(R
1
+ R
2
)
13.
?C = 2 ?F
? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other.
(a) ? The equation of capacitance in one row
C = 
3
C
(b) and three capacitance of capacity 
3
C
are connected in parallel
? The equation of capacitance 
C = 
3
C
3
C
3
C
? ? = C = 2 ?F 
As the volt capacitance on each row are same and the individual is 
= 
ce tan capaci of . No
Total
= 
3
60
= 20 V
14. Let there are ‘x’ no of capacitors in series ie in a row
So, x × 50 = 200 
? x = 4 capacitors.
Effective capacitance in a row = 
4
10
Now, let there are ‘y’ such rows,
So, 
4
10
× y = 10 
? y = 4 capacitor.
So, the combinations of four rows each of 4 capacitors.
6 ?F ?
B
5 ?F ?
10 V
A
A5 
B6 
C
A
B
C C
C C
C
C C C
Page 4


31.1
CHAPTER – 31
CAPACITOR
1. Given that
Number of electron = 1 × 10
12
Net charge Q = 1 × 10
12
× 1.6 × 10
–19
= 1.6 × 10
–7
C
? The net potential difference = 10 L.
? Capacitance – C = 
v
q
= 
10
10 6 . 1
7 ?
?
= 1.6 × 10
–8
F.
2. A = ?r
2
= 25 ?cm
2
d = 0.1 cm
c = 
d
A
0
?
= 
1 . 0
14 . 3 25 10 854 . 8
12
? ? ?
?
= 6.95 × 10
–5
?F.   
3. Let the radius of the disc = R
?Area = ?R
2
C = 1 ?
D = 1 mm = 10
–3
m
? C = 
d
A
0
?
? 1 = 
3
2 12
10
r 10 85 . 8
?
?
? ? ?
? r
2
= 
? ?
?
?
85 . 8
10 10
12 3
= 
784 . 27
10
9
= 5998.5 m = 6 Km
4. A = 25 cm
2
  = 2.5 × 10
–3
cm
2
d = 1 mm = 0.01 m
V = 6V Q = ?
C = 
d
A
0
?
= 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
Q = CV = 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
× 6 = 1.32810 × 10
–10
C
W = Q × V = 1.32810 × 10
–10
× 6 = 8 × 10
–10
J. 
5. Plate area A = 25 cm
2
  = 2.5 × 10
–3
m
Separation d = 2 mm = 2 × 10
–3
m
Potential v = 12 v
(a) We know C = 
d
A
0
?
  =  
3
3 12
10 2
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 11.06 ×10
–12
F
C = 
v
q
? 11.06 ×10
–12
= 
12
q
? q
1
= 1.32 × 10
–10
C. 
(b) Then d = decreased to 1 mm 
? d = 1 mm = 1 × 10
–3 
m
C = 
d
A
0
?
  = 
v
q
= 
3
3 12
10 1
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 
12
2
? q
2
= 8.85 × 2.5 × 12 × 10
–12
= 2.65 × 10
–10
C.
? The extra charge given to plate = (2.65 – 1.32) × 10
–10
= 1.33 × 10
–10
C.
6. C
1
= 2 ?F, C
2
= 4 ?F , 
C
3
= 6 ?F V = 12 V
cq = C
1
+ C
2
+ C
3
= 2 + 4 + 6 = 12 ?F = 12 × 10
–6
F
q
1
= 12 × 2 = 24 ?C, q
2
= 12 × 4 = 48 ?C, q
3
= 12 × 6 = 72 ?C 
5 cm
0.1 cm
1 mm
C 1 V C 2
C 3
Capacitor
31.2
7.
? The equivalent capacity.
C = 
2 1 3 1 3 2
3 2 1
C C C C C C
C C C
? ?
= 
30 20 40 20 40 30
40 30 20
? ? ? ? ?
? ?
= 
2600
24000
= 9.23 ?F
(a) Let Equivalent charge at the capacitor = q
C = 
V
q
? q = C × V = 9.23 × 12 = 110 ?C on each.
As this is a series combination, the charge on each capacitor is same as the equivalent charge 
which is 110 ?C. ?
(b) Let the work done by the battery = W
? V = 
q
W
? W = Vq = 110 × 12 × 10
–6
= 1.33 × 10
–3
J.
8. C
1
= 8 ?F, C
2
= 4 ?F , C
3
= 4 ?F
Ceq = 
3 2 1
1 3 2
C C C
C ) C C (
? ?
? ?
= 
16
8 8 ?
= 4 ?F
Since B & C are parallel & are in series with A
So, q
1
= 8 × 6 = 48 ?C q
2
= 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 
9. (a)
? C
1
, C
1
are series & C
2
, C
2
are series as the V is same at p & q. So no current pass through p & q.
2 1
C
1
C
1
C
1
? ? ?
2 1
C C
1 1
C
1 ?
?
C
p
= 
2
C
1
=
2
4
= 2 ?F
And C
q
= 
2
C
2
= 
2
6
= 3 ?F
? C = C
p
+ C
q
= 2 + 3 = 5 ?F
(b) C
1
= 4 ?F, C
2
= 6 ?F,
In case of p & q, q = 0
? C
p 
= 
2
C
1
= 
2
4
= 2 ?F
C
q
= 
2
C
2
= 
2
6
= 3 ?F
& C ? = 2 + 3 = 5 ?F
C & C ? = 5 ?F
? The equation of capacitor  C = C ? + C ?? = 5 + 5 = 10 ?F
12 V
8 ?F ?
A
4 ?F ?
B C
4 ?F ?
A
B
C 1
C 2
C 1
C 2
C 1 = 4
C 2 = 6
A
p
B
C 1
C 1
C 1 C 1
q
R
S
C 2
C 2
C 2
C 2
30 ?F ?
V = 12 V
40 ?F ? 20 ?F ?
Capacitor
31.3
10. V = 10 v 
Ceq = C
1
+C
2
[ ? They are parallel]
= 5 + 6 = 11 ?F
q = CV = 11 × 10 110 ?C ?
11. The capacitance of the outer sphere = 2.2 ?F
C = 2.2 ?F
Potential, V = 10 v
Let the charge given to individual cylinder = q.
C = 
V
q
? q = CV = 2.2 × 10 = 22 ?F
? The total charge given to the inner cylinder = 22 + 22 = 44 ?F
12. C = 
V
q
, Now V = 
R
Kq
So, C
1
= 
? ?
1
R / Kq
q
= 
K
R
1
= 4 ??
0
R
1
Similarly c
2
= 4 ??
0
R
2
The combination is necessarily parallel.
Hence Ceq = 4 ??
0
R
1
+4 ??
0
R
2
= 4 ??
0
(R
1
+ R
2
)
13.
?C = 2 ?F
? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other.
(a) ? The equation of capacitance in one row
C = 
3
C
(b) and three capacitance of capacity 
3
C
are connected in parallel
? The equation of capacitance 
C = 
3
C
3
C
3
C
? ? = C = 2 ?F 
As the volt capacitance on each row are same and the individual is 
= 
ce tan capaci of . No
Total
= 
3
60
= 20 V
14. Let there are ‘x’ no of capacitors in series ie in a row
So, x × 50 = 200 
? x = 4 capacitors.
Effective capacitance in a row = 
4
10
Now, let there are ‘y’ such rows,
So, 
4
10
× y = 10 
? y = 4 capacitor.
So, the combinations of four rows each of 4 capacitors.
6 ?F ?
B
5 ?F ?
10 V
A
A5 
B6 
C
A
B
C C
C C
C
C C C
Capacitor
31.4
15.
(a) Capacitor = 
8 4
8 4
?
?
= 
3
8
??
and 
3 6
3 6
?
?
= 2 ?F
(i) The charge on the capacitance 
3
8
?F 
? Q = 
3
8
× 50 = 
3
400
?
? The potential at 4 ?F = 
4 3
400
?
= 
3
100
at 8 ?F = 
8 3
400
?
= 
6
100
The Potential difference = 
6
100
3
100
? = 
3
50
?V
(ii) Hence the effective charge at 2 ?F = 50 × 2 = 100 ?F
? Potential at 3 ?F = 
3
100
;  Potential at 6 ?F = 
6
100
?Difference = 
6
100
3
100
? = 
3
50
?V
? The potential at C & D is 
3
50
?V
(b) ?
S
R
q
P
? = 
2
1
= 
2
1
= It is balanced. So from it is cleared that the wheat star bridge balanced. So 
the potential at the point C & D are same. So no current flow through the point C & D. So if we connect 
another capacitor at the point C & D the charge on the capacitor is zero.
16. Ceq between a & b
= 
2 1
2 1
3
2 1
2 1
C C
C C
C
C C
C C
?
? ?
?
= 
2 1
2 1
3
C C
C C 2
C
?
? ( ?The three are parallel)
17. In the figure the three capacitors are arranged in parallel.
All have same surface area = a = 
3
A
First capacitance C
1
= 
d 3
A
0
?
2
nd
capacitance C
2
= 
) d b ( 3
A
0
?
?
3
rd
capacitance C
3
= 
) d b 2 ( 3
A
0
?
?
Ceq = C
1
+ C
2
+C
3 
8 ?F ?
B
C
A
4 ?F ?
3 ?F ? 6 ?F ?
D
50
A
C
B
D
6 ?F ?
8 ?F ?
3 ?F ?
4 ?F ?
8 ?F ? C
D
50
4 ?F ?
3 ?F ? 6 ?F ?
C 1
C 2
b
a C 3
C 2
C 1
C 1C 2/C 1+C 2
b
a
C 3
C 1C 2/C 1+C 2
b
C
D B
a
A
d
a
a
b
E
Page 5


31.1
CHAPTER – 31
CAPACITOR
1. Given that
Number of electron = 1 × 10
12
Net charge Q = 1 × 10
12
× 1.6 × 10
–19
= 1.6 × 10
–7
C
? The net potential difference = 10 L.
? Capacitance – C = 
v
q
= 
10
10 6 . 1
7 ?
?
= 1.6 × 10
–8
F.
2. A = ?r
2
= 25 ?cm
2
d = 0.1 cm
c = 
d
A
0
?
= 
1 . 0
14 . 3 25 10 854 . 8
12
? ? ?
?
= 6.95 × 10
–5
?F.   
3. Let the radius of the disc = R
?Area = ?R
2
C = 1 ?
D = 1 mm = 10
–3
m
? C = 
d
A
0
?
? 1 = 
3
2 12
10
r 10 85 . 8
?
?
? ? ?
? r
2
= 
? ?
?
?
85 . 8
10 10
12 3
= 
784 . 27
10
9
= 5998.5 m = 6 Km
4. A = 25 cm
2
  = 2.5 × 10
–3
cm
2
d = 1 mm = 0.01 m
V = 6V Q = ?
C = 
d
A
0
?
= 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
Q = CV = 
01 . 0
10 5 . 2 10 854 . 8
3 12 ? ?
? ? ?
× 6 = 1.32810 × 10
–10
C
W = Q × V = 1.32810 × 10
–10
× 6 = 8 × 10
–10
J. 
5. Plate area A = 25 cm
2
  = 2.5 × 10
–3
m
Separation d = 2 mm = 2 × 10
–3
m
Potential v = 12 v
(a) We know C = 
d
A
0
?
  =  
3
3 12
10 2
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 11.06 ×10
–12
F
C = 
v
q
? 11.06 ×10
–12
= 
12
q
? q
1
= 1.32 × 10
–10
C. 
(b) Then d = decreased to 1 mm 
? d = 1 mm = 1 × 10
–3 
m
C = 
d
A
0
?
  = 
v
q
= 
3
3 12
10 1
10 5 . 2 10 85 . 8
?
? ?
?
? ? ?
= 
12
2
? q
2
= 8.85 × 2.5 × 12 × 10
–12
= 2.65 × 10
–10
C.
? The extra charge given to plate = (2.65 – 1.32) × 10
–10
= 1.33 × 10
–10
C.
6. C
1
= 2 ?F, C
2
= 4 ?F , 
C
3
= 6 ?F V = 12 V
cq = C
1
+ C
2
+ C
3
= 2 + 4 + 6 = 12 ?F = 12 × 10
–6
F
q
1
= 12 × 2 = 24 ?C, q
2
= 12 × 4 = 48 ?C, q
3
= 12 × 6 = 72 ?C 
5 cm
0.1 cm
1 mm
C 1 V C 2
C 3
Capacitor
31.2
7.
? The equivalent capacity.
C = 
2 1 3 1 3 2
3 2 1
C C C C C C
C C C
? ?
= 
30 20 40 20 40 30
40 30 20
? ? ? ? ?
? ?
= 
2600
24000
= 9.23 ?F
(a) Let Equivalent charge at the capacitor = q
C = 
V
q
? q = C × V = 9.23 × 12 = 110 ?C on each.
As this is a series combination, the charge on each capacitor is same as the equivalent charge 
which is 110 ?C. ?
(b) Let the work done by the battery = W
? V = 
q
W
? W = Vq = 110 × 12 × 10
–6
= 1.33 × 10
–3
J.
8. C
1
= 8 ?F, C
2
= 4 ?F , C
3
= 4 ?F
Ceq = 
3 2 1
1 3 2
C C C
C ) C C (
? ?
? ?
= 
16
8 8 ?
= 4 ?F
Since B & C are parallel & are in series with A
So, q
1
= 8 × 6 = 48 ?C q
2
= 4 × 6 = 24 ?C q3 = 4 × 6 = 24 ?C 
9. (a)
? C
1
, C
1
are series & C
2
, C
2
are series as the V is same at p & q. So no current pass through p & q.
2 1
C
1
C
1
C
1
? ? ?
2 1
C C
1 1
C
1 ?
?
C
p
= 
2
C
1
=
2
4
= 2 ?F
And C
q
= 
2
C
2
= 
2
6
= 3 ?F
? C = C
p
+ C
q
= 2 + 3 = 5 ?F
(b) C
1
= 4 ?F, C
2
= 6 ?F,
In case of p & q, q = 0
? C
p 
= 
2
C
1
= 
2
4
= 2 ?F
C
q
= 
2
C
2
= 
2
6
= 3 ?F
& C ? = 2 + 3 = 5 ?F
C & C ? = 5 ?F
? The equation of capacitor  C = C ? + C ?? = 5 + 5 = 10 ?F
12 V
8 ?F ?
A
4 ?F ?
B C
4 ?F ?
A
B
C 1
C 2
C 1
C 2
C 1 = 4
C 2 = 6
A
p
B
C 1
C 1
C 1 C 1
q
R
S
C 2
C 2
C 2
C 2
30 ?F ?
V = 12 V
40 ?F ? 20 ?F ?
Capacitor
31.3
10. V = 10 v 
Ceq = C
1
+C
2
[ ? They are parallel]
= 5 + 6 = 11 ?F
q = CV = 11 × 10 110 ?C ?
11. The capacitance of the outer sphere = 2.2 ?F
C = 2.2 ?F
Potential, V = 10 v
Let the charge given to individual cylinder = q.
C = 
V
q
? q = CV = 2.2 × 10 = 22 ?F
? The total charge given to the inner cylinder = 22 + 22 = 44 ?F
12. C = 
V
q
, Now V = 
R
Kq
So, C
1
= 
? ?
1
R / Kq
q
= 
K
R
1
= 4 ??
0
R
1
Similarly c
2
= 4 ??
0
R
2
The combination is necessarily parallel.
Hence Ceq = 4 ??
0
R
1
+4 ??
0
R
2
= 4 ??
0
(R
1
+ R
2
)
13.
?C = 2 ?F
? In this system the capacitance are arranged in series. Then the capacitance is parallel to each other.
(a) ? The equation of capacitance in one row
C = 
3
C
(b) and three capacitance of capacity 
3
C
are connected in parallel
? The equation of capacitance 
C = 
3
C
3
C
3
C
? ? = C = 2 ?F 
As the volt capacitance on each row are same and the individual is 
= 
ce tan capaci of . No
Total
= 
3
60
= 20 V
14. Let there are ‘x’ no of capacitors in series ie in a row
So, x × 50 = 200 
? x = 4 capacitors.
Effective capacitance in a row = 
4
10
Now, let there are ‘y’ such rows,
So, 
4
10
× y = 10 
? y = 4 capacitor.
So, the combinations of four rows each of 4 capacitors.
6 ?F ?
B
5 ?F ?
10 V
A
A5 
B6 
C
A
B
C C
C C
C
C C C
Capacitor
31.4
15.
(a) Capacitor = 
8 4
8 4
?
?
= 
3
8
??
and 
3 6
3 6
?
?
= 2 ?F
(i) The charge on the capacitance 
3
8
?F 
? Q = 
3
8
× 50 = 
3
400
?
? The potential at 4 ?F = 
4 3
400
?
= 
3
100
at 8 ?F = 
8 3
400
?
= 
6
100
The Potential difference = 
6
100
3
100
? = 
3
50
?V
(ii) Hence the effective charge at 2 ?F = 50 × 2 = 100 ?F
? Potential at 3 ?F = 
3
100
;  Potential at 6 ?F = 
6
100
?Difference = 
6
100
3
100
? = 
3
50
?V
? The potential at C & D is 
3
50
?V
(b) ?
S
R
q
P
? = 
2
1
= 
2
1
= It is balanced. So from it is cleared that the wheat star bridge balanced. So 
the potential at the point C & D are same. So no current flow through the point C & D. So if we connect 
another capacitor at the point C & D the charge on the capacitor is zero.
16. Ceq between a & b
= 
2 1
2 1
3
2 1
2 1
C C
C C
C
C C
C C
?
? ?
?
= 
2 1
2 1
3
C C
C C 2
C
?
? ( ?The three are parallel)
17. In the figure the three capacitors are arranged in parallel.
All have same surface area = a = 
3
A
First capacitance C
1
= 
d 3
A
0
?
2
nd
capacitance C
2
= 
) d b ( 3
A
0
?
?
3
rd
capacitance C
3
= 
) d b 2 ( 3
A
0
?
?
Ceq = C
1
+ C
2
+C
3 
8 ?F ?
B
C
A
4 ?F ?
3 ?F ? 6 ?F ?
D
50
A
C
B
D
6 ?F ?
8 ?F ?
3 ?F ?
4 ?F ?
8 ?F ? C
D
50
4 ?F ?
3 ?F ? 6 ?F ?
C 1
C 2
b
a C 3
C 2
C 1
C 1C 2/C 1+C 2
b
a
C 3
C 1C 2/C 1+C 2
b
C
D B
a
A
d
a
a
b
E
Capacitor
31.5
= 
d 3
A
0
?
+ 
) d b ( 3
A
0
?
?
+ 
) d b 2 ( 3
A
0
?
?
= ?
?
?
?
?
?
?
?
?
?
?
d b 2
1
d b
1
d
1
3
A
0
= 
?
?
?
?
?
?
?
?
? ?
? ? ? ? ? ? ?
) d b 2 )( d b ( d
d ) d b ( d ) d b 2 ( ) d b 2 )( d b (
3
A
0
= 
? ?
) d b 2 )( d b ( d 3
b 2 bd 6 d 3 A
2 2
0
? ?
? ? ?
18. (a) C = 
) R / R ( In
L 2
1 2
0
?
= 
2 In
10 10 85 . 8 14 . 3 e
1 2 ? ?
? ? ? ?
[In2 = 0.6932]
= 80.17 × 10
–13
? 8 PF
(b) Same as R
2
/R
1
will be same.
19. Given that
C = 100 PF = 100 × 10
–12
F C
cq
= 20 PF = 20 × 10
–12
F
V = 24 V q = 24 × 100 × 10
–12
= 24 × 10
–10
q
2
= ?
Let q
1
= The new charge 100 PF V
1
= The Voltage. 
Let the new potential is V
1
After the flow of charge, potential is same in the two capacitor
V
1
= 
2
2
C
q
= 
1
1
C
q
= 
2
1
C
q q ?
= 
1
1
C
q
= 
12
1
10
10 24
q 10 24
?
?
?
? ?
= 
12
1
10 100
q
?
?
= 24 × 10
–10 
– q
1 =
5
q
1
= 6q
1
= 120 × 10
–10
= q
1
= 
6
120
×10
–10 
= 20 × 10
–10
? V
1 = 
1
1
C
q
= 
12
10
10 100
10 20
?
?
?
?
= 20 V
20.
Initially when ‘s’ is not connected,
C
eff
= q
3
C 2
= 50
3
C 2
? = 
4
10
2
5
?
? = 1.66 × 10
–4
C
After the switch is made on,
Then C
eff
= 2C = 10
–5
Q = 10
–5
× 50 = 5 × 10
–4
Now, the initial charge will remain stored in the stored in the short capacitor 
Hence net charge flowing
= 5 × 10
–4
– 1.66 × 10
–4
= 3.3 × 10
–4
C. 
A
S
/
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