Question

Half of the space between the plates of a parallel plate capacitor is filled with a medium of dielectric constant $K$ parallel to the plates. If initially the capacitance is $C$, then the new capacitance will be

• A. $\frac{KC}{K+1}$
• B. $KC$
• C. $\frac{(K+1)C}{2}$
• D. $\frac{2KC}{K+1}$

Answer 1

The correct option is D $\frac{2KC}{K+1}$

Let us assume that the area of the plates is $A$ and the separation between them is $d$.

So, initial capacitance will be

$C=\frac{{ϵ}_{0}A}{d}$

The capacitance with the dielectric of thickness $t(=\frac{d}{2})$ will be

$C^{\prime }=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}=\frac{{ϵ}_{0}A}{d-\frac{d}{2}+\frac{d}{2K}}$

$\Rightarrow C^{\prime }=\frac{2{ϵ}_{0}AK}{d(K+1)}$

$\Rightarrow C^{\prime }=\frac{2KC}{K+1}$

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

Key concept: Capacitance with partial dielectric is given by $C=\frac{{ϵ}_{0}A}{d-t+\frac{t}{K}}$