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WBJEE 2014 Physics Solved Paper
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© examsnet.com
Question : 60
Total: 60
Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectricconstant K. The remaining half contains air as shown in the figure. The capacitor is now given a charge Q. Then
electric field in the dielectric-filled region is higher than that in the air-filled region
on the two halves of the bottom plate the charge densities are unequal
charge on the half of the top plate above the air-filled part is
Q
K
+
1
capacitance of the capacitor shown above is (1 + K)
C
0
2
,
C
0
is the capacitance of the same capacitor withthe dielectric removed
Validate
Solution:
C
1
=
K
∊
0
A
2
d
,
C
2
=
∊
0
A
2
d
C
eq
=
∊
A
2
d
(K + 1) =
C
0
2
(K + 1),
Q
1
Q
2
=
C
1
C
2
=
K
1
⇒
σ
1
σ
2
=
K
1
Q
1
=
K
Q
K
+
1
and
Q
2
=
Q
K
+
1
,
E =
σ
∊
0
K
,
E
1
E
2
=
σ
1
σ
2
×
K
2
K
1
=
Q
1
Q
2
×
K
2
K
1
=
K
1
×
1
K
= 1 : 1
© examsnet.com
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