Question

The initial capacitance of a parallel plate capacitor is $C_0$ and after introducing a conducting sheet of thickness $\frac{d}{3}$, the capacitance becomes $C$. Then $\frac{C}{C_0}$ is

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

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

Capacitance without dielectric is given by

$C_0=\frac{{ϵ}_{0}A}{d}$

As we know that capacitance of partially filled dielectric slab of thickness $t$ and dielectric constant $K$ is given by

$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$

Given:
Thickness of dielectric, $t=\frac{d}{3}$
Dielectric constant of a conductor,$K=\infty$

Hence, new capacitance will be,

$C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}=\frac{{ϵ}_{0}A}{d-\frac{d}{3}+\frac{d}{3\times \infty }}$

$\Rightarrow C=\frac{3{ϵ}_{0}A}{2d}=\frac{3}{2}{C}_0$

$\therefore \frac{C}{C_0}=\frac{3}{2}$

Hence, option $\left(b\right)$ is correct.

Key concept- Dielectric constant of a metal is infinite.