Question

Two identical metal plates separated by a distance $d$ forms parallel plate capacitor of capacity $C$. A metal sheet of thickness $d/2$ and same dimensions is inserted between the plates, so that the air gap is separated into two equal parts. The new capacity of the system will be:

• A. $\frac{{\epsilon }_{0}A}{2d}$
• B. $\frac{{\epsilon }_{0}A}{4d}$
• C. $\frac{2{ϵ}_{0}A}{d}$
• D. $\frac{{ϵ}_{0}A}{d}$

Answer 1

Answer: (C)

The final capacitance in series combination of
$\frac{{\epsilon }_{0}A}{\frac{d}{4}}$ and $\frac{{\epsilon }_{0}A}{\frac{d}{4}}$ is
$\frac{1}{C}=\frac{2d}{4{ϵ}_{0}A}$
Hence , $C=\frac{2{ϵ}_{0}A}{d}$