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 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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FAQs on HC Verma Solutions: Chapter 31 - Capacitors - Physics Class 11 - NEET

1. What is a capacitor and how does it work?
Ans. A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, positive and negative charges accumulate on each plate, creating an electric field between them. This electric field stores the electrical energy.
2. How is the capacitance of a capacitor determined?
Ans. The capacitance of a capacitor is determined by its physical characteristics, specifically the surface area of the plates, the distance between them, and the type of dielectric material used. The capacitance is directly proportional to the surface area of the plates and the permittivity of the dielectric material, and inversely proportional to the distance between the plates. It is usually measured in farads (F).
3. What is the role of a dielectric material in a capacitor?
Ans. The dielectric material in a capacitor serves as an insulator between the two plates, preventing direct contact and the flow of current between them. It also increases the capacitance of the capacitor by reducing the electric field strength between the plates. Different dielectric materials have different permittivity values, which affect the capacitance and other properties of the capacitor.
4. Can a capacitor store an unlimited amount of charge?
Ans. No, a capacitor has a maximum charge that it can store, determined by its capacitance value. The maximum charge a capacitor can store is directly proportional to its capacitance. Once the capacitor is charged to its maximum capacity, it cannot store any more charge unless the existing charge is discharged or removed.
5. What are some common applications of capacitors?
Ans. Capacitors have various applications in electronic circuits. Some common applications include energy storage in power supplies, smoothing out voltage fluctuations, filtering out noise and interference, timing circuits, motor starting and running, tuning and filtering in radio frequency circuits, and energy storage in flash cameras. Capacitors also play a crucial role in various electronic devices, such as televisions, computers, and mobile phones.
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