If a thin metal foil of same area is placed between the two plates of a parallel plate capacitor of capacitance $C$ , then new capacitance will be :

C

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2 C

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3 C

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4 C

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Solution

The correct option is A $C$
We have

$C = \frac{ε_{0} A}{d}$

After placement of plate of equal area in between two plates we have two capacitors in series.

Value of each capacitor

$C_{1} = \frac{ε_{0} A}{d / 2}$

$C_{1} = \frac{2 ε_{0} A}{d} = 2 C$

Equivalent capacitance

$C_{e q} = \frac{2 C \times 2 C}{2 C + 2 C}$

$C_{e q} = C$

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