Question

Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor, are placed. If the thickness of these plates is equal to ${\left\{\frac{1}{5}\right\}}^{th}$ of the distance between the plates of the original capacitor, then the capacity of the new capacitor is :

• A. $\frac{5}{3}C$
• B. $\frac{3}{5}C$
• C. $\frac{3}{10}C$
• D. $\left(\frac{10}{3}\right)C$

Answer 1

The correct option is A $\frac{5}{3}C$

Initial capacitance
$C=\frac{{\epsilon }_{0}A}{d}$

Now assume the plates are arranged as follows.

The above arrangement acts as three capacitors kept in series.

$C_1=\frac{{\epsilon }_{0}A}{x_1},C_2=\frac{{\epsilon }_{0}A}{x_2},C_3=\frac{{\epsilon }_{0}A}{x_3}$

$\frac{1}{C_{eff}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\Rightarrow \frac{1}{C_{eff}}=\frac{x_1+x_2+x_3}{{\epsilon }_{0}A}$

$\Rightarrow C_{eff}=\frac{{\epsilon }_{0}A}{x_1+x_2+x_3}$

$x_1+x_2+x_3=\frac{3d}{5}$

$\therefore C_{eff}=\frac{5{\epsilon }_{0}A}{3d}=\frac{5C}{3.}$