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

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

The correct option is A CWe have,C=ϵ0AdAfter placement of plate of equal area in between the two plates, we have two capacitors in series.Value of each capacitor,C1=ϵ0Ad/2C1=2ϵ0Ad=2CEquivalent capacitance, Ceq=2C×2C2C+2C⟹Ceq=C