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JEE Main Physics Class 12 Electrostatic Potential and Capacitance Part 3 Questions
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© examsnet.com
Question : 18
Total: 57
A capacitor of
10
µ
F
capacitance whose plates are separated by
10
mm
through air and each plate has area
4
cm
2
is now filled equally with two dielectric media of
K
1
=
2
,
K
2
=
3
respectively as shown in figure. If new force between the plates is
8
N
. The supply voltage is ______ V.
[6 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
👈: Video Solution
C
eq
=
C
1
+
C
2
C
1
=
2
E
0
A
2
×
d
=
10
µ
F
C
2
=
3
E
0
A
2
d
=
15
µ
F
C
eq
=
25
µ
F
Now the charge on
C
1
=
10
V
µ
c
C
2
=
1.5
V
µ
C
.
Now force between the plates
[
F
=
Q
2
2
A
∈
0
]
100
V
2
×
10
−
12
2
×
2
×
10
−
4
E
0
+
225
V
2
×
10
−
12
2
×
2
×
10
−
4
×
E
0
=
8
325
V
2
=
8
×
4
×
10
−
4
×
8.85
V
2
=
32
×
8.85
×
10
−
4
325
∴
V
=
√
283.2
×
10
−
4
325
V
=
0.93
×
10
−
2
© examsnet.com
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