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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 1 Questions
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Question : 10
Total: 100
A 60 pF capacitor is fully charged by a 20 V supply. It is then disconnected from the supply and is connected to another uncharged 60 pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ)
[NA 7 Jan. 2020 II]
Your Answer:
Validate
Solution:
In the first condition, electrostatic energy is
U
i
=
1
2
C
V
0
2
=
1
2
×
60
×
10
−
12
×
400
=
12
×
10
−
9
J
In the second condition
U
F
=
1
2
C
′
V
′
2
U
f
=
1
2
2
C
.
(
V
0
2
)
2
(
∵
C
′
=
2
C
,
V
′
=
V
0
2
)
=
1
4
×
60
×
10
−
12
×
(
20
)
2
=
6
×
10
−
9
J
Energy lost
=
U
i
−
U
f
=
12
×
10
−
9
J
−
6
×
10
−
9
J
=
6
n
J
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