Question

The capacitance of a capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t=d/2$ is inserted between the plates ($d$ is the separation between the plates). What is the dielectric constant of the slab?

Answer 1

The capacitance without dielectric is $C=\frac{A{ϵ}_0}{d}.$
When the dielectric slab of thickness $t=d/2$ and dielectric constant $k$ is inserted between the plates, the capacitance becomes
$C^{\prime }=\frac{A{ϵ}_0}{d-t(1-\frac{1}{k})}=\frac{A{ϵ}_0}{(d-\frac{d}{2}+\frac{d}{2k})}$
now , $C^{\prime }=\left(4/3\right)C\Rightarrow \frac{A{ϵ}_0}{(d-\frac{d}{2}+\frac{d}{2k})}=\left(4/3\right)\frac{A{ϵ}_0}{d}$
or $\frac{1}{\frac{1}{2}+\frac{1}{2k}}=\frac{4}{3}$
or $2+\frac{2}{k}=3\Rightarrow k=2$