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Question

Two identical sheets of the same metallic foil of the area A are separated by a distance d, and charged by a battery of emf E. Keeping the charge constant, the separation is increased by l. Then, the new capacitance and potential difference will be

A
ϵ0Ad, [1+ld]E
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B
ϵ0A(d+l), [1+ld]E
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C
ϵ0Ad, E
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D
ϵ0A(d+l), E
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Solution

The correct option is B ϵ0A(d+l), [1+ld]E
Initial capcitance C will be

C=Aϵ0d

If charge is kept constant,

Qi=Qf .......(1)

By using formula, Q=CV

Qi=Aϵ0dE

Let the final potential difference be E.

Final capacitance C when separation is increased by l will be

C=Aϵ0d+l

Again using formula, Q=CV

Qf=Aϵ0d+lE

putting the values in eq.(1)

(Aϵ0d+l)E=Aε0dE

E=E(1+ld)

Hence, option (c) is the correct choice.

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