Question

A slab of dielectric constant $K$ has the same crosssectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$, where $d$ is thenseparation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:
(Given $C_0=$capacitance of capacitor with air as medium between plates.)

Answer 1

Given:
dielectric constant $=K$
thickness $=\frac{3}{4}d$,
Capacitance,
$C_0=\frac{{\epsilon }_{0}A}{d}$
when the slab is inserted,
$C=\frac{{\epsilon }_{0}A}{d-\frac{3d}{4}+\frac{3d}{4K}}=\frac{4{\epsilon }_{0}AK}{3d+Kd}$

$C=\frac{4K}{3+K}\left(\frac{{\epsilon }_{0}A}{d}\right)=\frac{4K{C}_0}{3+K}$

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