Question

A parallel plate capacitor of plate separation d, plate area A and of capacitance C is charged to a potential difference V by connecting it to a battery. When a dielectric slab of thickness t and k = 2 is introduced between the plates of the capacitor, its capacitance increases to $\frac{3}{2}C$, the ratio $\frac{t}{d}$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A $\frac{2}{3}$

For air capacitor, $C_1=\frac{{ϵ}_{0}A}{d}$
For capacitor with dielectric, $C_2=\frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}$
But $C_2=\frac{3}{2}{C}_1\Rightarrow \frac{{ϵ}_{0}A}{d-t(1-\frac{1}{k})}=\frac{3}{2}\frac{{ϵ}_{0}A}{d}$
$\therefore d-t(1-\frac{1}{k})=\frac{2}{3}d\Rightarrow \frac{d}{3}=t(1-\frac{1}{k})=t(1-\frac{1}{2})=\frac{t}{2}$
$\therefore \frac{t}{d}=\frac{2}{3}$