Question

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 plates (d = separation between the plates). Then the dielectric constant of the slab is

Answer 1

From given $C_f=\frac{4}{3}{C}_i$
After inserting dielectric the capacitor can be redrawn as shown below

$C_1=\frac{{\epsilon }_{0}A}{d/2};C_2=\frac{K{\epsilon }_{0}A}{d/2}$
Now,
$C_{eq}=\frac{C_1\times C_2}{C_1+C_2}$
On solving we get
$C_{eq}=\frac{2K{\epsilon }_{o}A}{d(K+1)}=C_f$
From given condition ,
$\frac{2K{\epsilon }_{0}A}{d(K+1)}=\frac{4}{3}\frac{{\epsilon }_{0}A}{d}$
And on solving the above equation, we get $K=2$.