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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 1 Questions
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© examsnet.com
Question : 45
Total: 100
A parallel plate capacitor with area
200
c
m
2
and separation between the plates
1.5
c
m
,
is connected across a battery of emfV. If the force of attraction between the plates is
25
×
10
−
6
N
,
the value of
V
is approximately.
(
ε
0
=
8.85
×
10
−
12
C
2
N
.
m
2
)
[Online April 15, 2018]
150V
100V
250V
300V
Validate
Solution:
Given area of Parallel plate capacitor,
A
=
200
c
m
2
Separation between the plates,
d
=
1.5
c
m
Force of attraction between the plates,
F
=
25
×
10
−
6
N
F
=
Q
E
F
=
Q
2
2
A
∈
0
(E due to parallel plate
=
σ
2
∈
0
=
Q
A
2
∈
0
)
But
Q
=
C
V
=
∈
0
A
(
V
)
d
∴
F
=
(
∈
0
A
V
)
2
d
2
×
2
A
∈
0
=
(
∈
0
A
)
2
×
V
2
d
2
×
2
×
(
A
∈
0
)
=
(
∈
0
A
)
×
V
2
d
2
×
2
or,
25
×
10
−
6
=
(
8.85
×
10
−
12
)
×
(
200
×
10
−
4
)
×
V
2
2.25
×
10
−
4
×
2
⇒
V
=
√
25
×
10
−
6
×
2.25
×
10
−
4
×
2
8.85
×
10
−
12
×
200
×
10
−
4
≈
250
V
© examsnet.com
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