All Courses
Get the App Signup / Login
Sign in Open App
NEET Exam  >  NEET Notes  >  Physics Class 11  >  HC Verma Solutions: Chapter 31 - Capacitors

HC Verma Solutions: Chapter 31 - Capacitors | Physics Class 11 - NEET PDF Download

Download, print and study this document offline
Please wait while the PDF view is loading
 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
/
Read More
102 videos|411 docs|121 tests
Join Course for Free

Top Courses for NEET

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.
Related Exams
About this Document
2.6K Views
4.97/5 Rating
Nov 15, 2024 Last updated
Document Description: HC Verma Solutions: Chapter 31 - Capacitors for NEET 2024 is part of Physics Class 11 preparation. The notes and questions for HC Verma Solutions: Chapter 31 - Capacitors have been prepared according to the NEET exam syllabus. Information about HC Verma Solutions: Chapter 31 - Capacitors covers topics like and HC Verma Solutions: Chapter 31 - Capacitors Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for HC Verma Solutions: Chapter 31 - Capacitors.
Introduction of HC Verma Solutions: Chapter 31 - Capacitors in English is available as part of our Physics Class 11 for NEET & HC Verma Solutions: Chapter 31 - Capacitors in Hindi for Physics Class 11 course. Download more important topics related with notes, lectures and mock test series for NEET Exam by signing up for free. NEET: HC Verma Solutions: Chapter 31 - Capacitors | Physics Class 11 - NEET
Description
Full syllabus notes, lecture & questions for HC Verma Solutions: Chapter 31 - Capacitors | Physics Class 11 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 11 | Best notes, free PDF download
Information about HC Verma Solutions: Chapter 31 - Capacitors
In this doc you can find the meaning of HC Verma Solutions: Chapter 31 - Capacitors defined & explained in the simplest way possible. Besides explaining types of HC Verma Solutions: Chapter 31 - Capacitors theory, EduRev gives you an ample number of questions to practice HC Verma Solutions: Chapter 31 - Capacitors tests, examples and also practice NEET tests
102 videos|411 docs|121 tests
Join Course for Free
Download as PDF
Explore Courses for NEET exam

Top Courses for NEET

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Related Searches

Objective type Questions

,

Viva Questions

,

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

,

Exam

,

shortcuts and tricks

,

Previous Year Questions with Solutions

,

Extra Questions

,

pdf

,

MCQs

,

Summary

,

mock tests for examination

,

Important questions

,

Semester Notes

,

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

,

Free

,

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

,

practice quizzes

,

video lectures

,

study material

,

past year papers

,

ppt

,

Sample Paper

;
Additional Information about HC Verma Solutions: Chapter 31 - Capacitors for NEET Preparation

HC Verma Solutions: Chapter 31 - Capacitors Free PDF Download

The HC Verma Solutions: Chapter 31 - Capacitors is an invaluable resource that delves deep into the core of the NEET exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective. With the help of these notes, you can grasp complex subjects quickly, revise important points easily, and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner, allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material, or toppers' notes, this PDF has got you covered. Download the HC Verma Solutions: Chapter 31 - Capacitors now and kickstart your journey towards success in the NEET exam.

Importance of HC Verma Solutions: Chapter 31 - Capacitors

The importance of HC Verma Solutions: Chapter 31 - Capacitors cannot be overstated, especially for NEET aspirants. This document holds the key to success in the NEET exam. It offers a detailed understanding of the concept, providing invaluable insights into the topic. By knowing the concepts well in advance, students can plan their preparation effectively. Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.

HC Verma Solutions: Chapter 31 - Capacitors Notes

HC Verma Solutions: Chapter 31 - Capacitors Notes offer in-depth insights into the specific topic to help you master it with ease. This comprehensive document covers all aspects related to HC Verma Solutions: Chapter 31 - Capacitors. It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation. Practice papers and question papers enable you to assess your progress effectively. Additionally, the paper analysis provides valuable tips for tackling the exam strategically. Access to Toppers' notes gives you an edge in understanding complex concepts. Whether you're a beginner or aiming for advanced proficiency, HC Verma Solutions: Chapter 31 - Capacitors Notes on EduRev are your ultimate resource for success.

HC Verma Solutions: Chapter 31 - Capacitors NEET Questions

The "HC Verma Solutions: Chapter 31 - Capacitors NEET Questions" guide is a valuable resource for all aspiring students preparing for the NEET exam. It focuses on providing a wide range of practice questions to help students gauge their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation. The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level. Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.

Study HC Verma Solutions: Chapter 31 - Capacitors on the App

Students of NEET can study HC Verma Solutions: Chapter 31 - Capacitors alongwith tests & analysis from the EduRev app, which will help them while preparing for their exam. Apart from the HC Verma Solutions: Chapter 31 - Capacitors, students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc. The EduRev App will make your learning easier as you can access it from anywhere you want. The content of HC Verma Solutions: Chapter 31 - Capacitors is prepared as per the latest NEET syllabus.
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup