Question

A slab of material of dielectric constant $k$ has the same area $A$ as the plates of a parallel plate capacitor and has a thickness $\left(3/4d\right)$, where $d$ is the separation of the plates. The change in capacitance when the slab is inserted between the plates is

• A. C=$\frac{{A\epsilon }_0}{d}\left(\frac{K+3}{4K}\right)$
• B. C=$\frac{{A\epsilon }_0}{d}\left(\frac{2K}{K+3}\right)$
• C. C=$\frac{{A\epsilon }_0}{d}\left(\frac{K+3}{2K}\right)$
• D. C=$\frac{{A\epsilon }_0}{d}\left(\frac{4K}{K+3}\right)$

Answer 1

The correct option is D C=$\frac{{A\epsilon }_0}{d}\left(\frac{4K}{K+3}\right)$

Here, thickness of the slab $t=\frac{3}{4}d$
Capacitance
$C=\frac{{\epsilon }_{0}A}{d-(t-\frac{t}{K})}=\frac{{\epsilon }_{0}A}{d-\frac{3}{4}d(1-\frac{1}{K})}=\frac{{\epsilon }_{0}A}{\frac{d}{4}+\frac{3}{4}\frac{d}{K}}=\frac{{\epsilon }_{0}A}{\frac{d}{4}(1+\frac{3}{K})}=\frac{{\epsilon }_{0}A}{d}\frac{4K}{(K+3)}$