Question

Potential difference across a parallel plate capacitor is $100.\text{V}$. Now a slab of $2\text{mm}$ thickness and dielectric constant ${ϵ}_r$ is inserted in it. If to maintain same potential difference distance between plates is increased by $1.6\text{mm}$. Find out ${ϵ}_r$ of slab

Answer 1

Given that
$C=\frac{Q}{V}=\text{Constant}\frac{{ϵ}_{0}A}{d}=\frac{{ϵ}_{0}A}{(d^{{}^{\prime }}-t+\frac{t}{{ϵ}_r})}$
$\Rightarrow d=d^{{}^{\prime }}-t+\frac{t}{{ϵ}_r}d=(d+1.6)-2+\frac{2}{{ϵ}_r}\Rightarrow {ϵ}_r=5.00$