Question

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has thickness $3d/4$, where d is the distance between plates. How is the capacitors changed when the slab is inserted between the plates?

Answer 1

$C=\frac{A{ϵ}_0}{d}$

$C^{\prime }=\frac{A{ϵ}_0}{d-t+\frac{t}{k}}$

Put $t=\frac{3d}{4}$ ;

$C^{\prime }=\frac{4k}{3+k}.\frac{A{ϵ}_0}{d}$

$C^{\prime }=\frac{4k}{3+k}.C$