Two metallic plates form a parallel plate capacitor. The distance between the plates is 'd '. A metal sheet of thickness $\frac{d}{2}$ and of area equal to area of each plate is introduced between the plates. What will be the ratio of the new capacitance to the original capacitance of the capacitor ?

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1:4

1:2

4:1

2:1

answer is A.

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Detailed Solution

$\begin{aligned} C_{1} = \frac{ϵ_{0} A}{d} \\ C_{2} = \frac{ϵ_{0} A}{\frac{d}{2} + \frac{d / 2}{\propto}} = \frac{2 \in_{0} A}{d} \\ \frac{C_{2}}{C_{1}} = \frac{2}{1} \end{aligned}$

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