The capacitance of a capacitor becomes $\frac{4}{3}$times its original value. If a dielectric slab of thickness $t=\frac{d}{2}$is inserted between the plate (d is the separation between the plates). The dielectric constant of the slab is

Let $E_{\circ },$$V_{\circ }$ and $Q_{\circ }$ be the initial electric field, potential and charge on the capacitor. 

Let dielectric slab of dielectric constant $K$ be inserted between the plates.

Then, total potential across the plates will be $V=E_0\frac{d}{2}+\frac{E_0}{K}\frac{d}{2}$

where $d$ is the separation between the plates.

$V=E_0\frac{d}{2}\left(\frac{K+1}{K}\right)=V_0\left(\frac{K+1}{2K}\right)\left(\because V_0=E_{0}d\right)$

New capacitance will be,$C=\frac{Q_0}{V}=\frac{2K}{K+1}\frac{Q_0}{V_0}=\frac{2K}{K+1}{C}_0$

According to question $C=\frac{4}{3}{C}_0$

$\therefore \frac{2K}{K+1}{C}_0=\frac{4}{3}{C}_0\text{ or }K=2$