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Question

The plates of a parallel plate capacitor are charged up to 100 V. Now, after removing the battery, a 2 mm thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. Dielectric constant of the plate is

A
5
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B
1.25
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C
4
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D
2.5
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Solution

The correct option is A 5
Initially,
and finally
and that can be redrawn as

As Q, V are constant C should also be constant.
Cf=C1×C2C1+C2=kε0A2×ε0A(d0.4)kε0A2+ε0A(d0.4)
Therfore,

ε0Ad=kε0A2×ε0A(d0.4)kε0A2+ε0A(d0.4)

On solving this, you get k=5.

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