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JEE Mains 05-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 2
Total: 75
A parallel plate capacitor has plate of length 'l', width 'w' and separation of plates is 'd'. It is connected to a battery of emf
V
. A dielectric slab of the same thickness 'd' and of dielectric constant
k
=
4
is being inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored?
l
/ 4
l
/ 2
l
/ 3
2
l
3
Validate
Solution:
Before inserting slab
C
i
=
ε
0
A
d
C
i
=
ε
0
l
w
d
After inserting dielectric slab
C
f
=
C
1
+
C
2
C
f
=
K
ε
0
A
1
d
+
ε
0
A
2
d
C
f
=
K
ε
0
w
x
d
+
ε
0
w
(
l
−
x
)
d
C
f
=
2
C
i
⇒
K
ε
0
w
x
x
+
ε
0
w
(
l
−
x
)
d
=
2
ε
0
l
w
d
4
x
+
l
−
x
=
2
l
[
x
=
l
3
]
© examsnet.com
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