Chapter 4 Capacitors
Your Turn
1
. If
Q
is doubled, the potential (
V
) will also double.
C
will not charge. It depends only on geometry of the
conductor.
2.
Q
=
CV
= (2µF)(1000 V)
= 2000 µC
= 2mC
Remember that farad X volt is coulomb.
3
. Solution:Potential of the two conductors is:
V
1
=
1
1
20
2
Q
C
=
= 10 V
V
2
=
2
2
40
4
Q
C
=
= 10 V
Since they have equal potentials, no charge will
flow between them when they are connected. No
heat is dissipated.
4.
If capacitance of B is
C
, then capacitance of A will
be 2
C
. After connection, the final common potential
of the two spheres will be
V
=
1
2
20
20
2
3
Q
Q
C
C
C
C
C
+
−
µ
+
µ
=
+
= 0
Hence, final charge on both the conductors will be
zero.
5
.
Final potential of the two spheres will be:
V
=
1
2
20
0
20
2
3
3
Q
Q
C
V
C
C
C
C
+
−
µ
+
=
=
−
µ
+
∴
q
A
= (2
C
)
V
=
40
3
C
−
µ
6.
E
=
V
d
=
3
50
1 10
−
×
= 50,000 V/m.
7
.
The electric field between the plates becomes
1
5
times. Therefore, the potential difference also
becomes
1
5
times.
8
. C =
0
A
d
∈
And 2C =
(
)
0
A
x
d
x
K
∈
+
−
2
0
A
d
∈
=
(
)
0
3
A
x
d
x
∈
+
−
2
3
x
+ 2(
d – x
) =
d
d
=
4
3
x
⇒
x
d
=
3
4
9
.
Refer to Example 6 for the reason explained
in Example 6; the capacitance of the spherical
capacitor will be
C =
0
4
K
ab
b
a
π
∈
−
=
9
3
0.2
0.1
9
10 [0.2
01]
×
×
×
−
= 6.7 × 10
–11
F.
10
.
Electric field between the plates is independent of
distance between them.
E
=
0
σ
∈
Therefore,
E
will not change when distance between
the plates is change.
∴
Energy density (
u
=
1
2
ε
0
E
2
) will not change
11
.
Half the work done by the cell is lost as heat and
remaining half gets stored in the capacitor.
12
.
W
cell
= 100μJ
Q
·
V = 100Μj
(5μC)
·
V
= 100μJ
V
= 20 V
13
.
Charge on the capacitor after it is connected to the
first cell is
Q
1
=
CV
Work done by the first cell is
W
1
=
Q
1
V
=
CV
2
Charge on the capacitor after it is connected to the
second cell is
Q
2
=
C
(2
V
) = 2
CV
Charge supplied by the second cell is =
Q
2
–
Q
1
=
CV
Work done by the second cell is
W
2
= (
CV
)(2V) = 2
CV
2
∴
2
1
2
2
1
2
2
W
CV
W
CV
=
=
Chapter_04 - Solution.indd
55
16-05-2019
19:30:22

S.56
Electrostatics and Current Electricity
14
.
Initial capacitance is
C
= 10µF
Final capacitance is 2C = 20µF
Initial charge is
Q
1
=
CV
(10µF)(20V)=200µC
Final charge is
Q
2
= 2
CV
= 400 µC
W
cell
= ∆
QV
= (
Q
2
–
Q
1
)
V
= (200µC)(20V) = 4000 µJ
= 4mJ
Change in energy stored in the capacitor is
∆
U
=
U
2
– U
1
=
1
2
(2C)
V
2
–
CV
2
=
1
2
CV
2
=
1
2
× (10µF)×(20V)
2
= 000 µJ = 2 mJ
W
extAgt
+
W
cell
= ∆
U
or,
W
extAgt
= 2 mJ – 4 mJ = –2mJ
Electric field between the plates E =
V
d
When
d
is halved, E doubles.
Energy density,
u
=
2
0
1
2
E
∈
If E doubles,
u
becomes 4 times.
15
. Equivalent capacitance is given by
0
1
1
1
1
1
2
3
C
=
+
+
C
0
=
6
11
F
µ
Charge on each capacitor is same as charge supplied
by the cell.
Q
=
C
0
V
=
6
11
F
µ
(66 V) = 36 µC
∴
Potential difference across the capacitors is
V
1
=
1
36
1
Q
C
=
= 36 V
V
2
=
2
36
2
Q
C
=
= 18 V
V
3
=
3
36
2
Q
C
C
F
µ
=
µ
= 12 V
16
.
When two capacitors are connected in parallel, the
equivalent capacitance is
C
0
=
C
1
+
C
2
.
When connected in series, the equivalent I less than
both
C
1
and
C
2
.
Therefore, parallel equivalent is 25µF and the series
equivalent is 4µF (the smallest one).
The two capacitances are 20µF and 5µF.
C
1
+
C
2
= 25µF
and
1
2
1
2
20
5
4
20
5
C C
F
C
C
×
=
=
µ
+
+
17
.
In both the diagrams, the three capacitors are in
parallel.

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