Question

For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4}d,$ where ${}^{\prime }{d}^{\prime }$ is the separation between the plates of the parallel plate capacitor. The new capacitance $\left(C^{\prime }\right)$ in terms of the original capacitance $\left(C_0\right)$ is given by the following relation:

• A. $C^{\prime }=\frac{4K}{K+3}{C}_0$
• B. $C^{\prime }=\frac{4}{3+K}{C}_0$
• C. $C^{\prime }=\frac{3+4}{4K}{C}_0$
• D. $C^{\prime }=\frac{4+K}{3}{C}_0$

Answer 1

The correct option is A $C^{\prime }=\frac{4K}{K+3}{C}_0$


Original capacitance
$C_0=\frac{{ϵ}_{0}A}{d}$
where $A$ is the area of the plates.

After inserting the dielectric, $C_1$ and $C_2$ are in series and $C^{\prime }$ is the new capacitance
$\therefore \frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_2}$
$\frac{1}{C^{\prime }}=\frac{\left(3d/4\right)}{{ϵ}_{0}KA}+\frac{\left(d/4\right)}{{ϵ}_{0}A}$
$\frac{1}{C^{\prime }}=\frac{d}{4{ϵ}_{0}A}\left(\frac{3+K}{K}\right)$

$\therefore C^{\prime }=\frac{4K}{(K+3)}{C}_0$
Hence, option (a) is correct.