Question

Capacitance of a parallel plate capacitor becomes $\frac{4}{3}$ times its original value if a dielectric slab of thickness $t=\frac{d}{2}$ is inserted between the plates [$d$ is the separation between the plates]. The dielectric constant of the slab is

• A. $4$
• B. $8$
• C. $2$
• D. $6$

Answer 1

The correct option is C $2$

Original value of capacitance be $C_o=\frac{{ϵ}_{0}A}{d}$
Let dielectric constant of the slab is $K$.

Now, Consider two capacitor $C_1\text{(with dielectric slab)}$ and $C_2\text{(without dielectric slab)}$ in series.
$C_1=\frac{K{ϵ}_{0}A}{\frac{d}{2}}=\frac{2K{ϵ}_{0}A}{d}$
$C_2=\frac{{ϵ}_{0}A}{\frac{d}{2}}=\frac{2{ϵ}_{0}A}{d}$
$\therefore C_{eq}=\frac{C_1{C}_2}{C_1+C_2}=\left(\frac{2K}{K+1}\right)\times \frac{{ϵ}_{0}A}{d}=\left(\frac{2K}{K+1}\right)\times C_o$

Given:
$C_{eq}=\frac{4}{3}{C}_o$
$\Rightarrow \frac{2K}{K+1}=\frac{4}{3}$
$\Rightarrow K=2$