The plates of a parallel plate capacitor are charged to $200V$ and then, the charging battery is disconnected. Now, a dielectric slab of dielectric constant $5$ and thickness $4mm$ is inserted between the capacitor plates. To maintain the original capacity, the increase in the separation between the plates of the capacitor is:

The correct option is B $3.2mm$
Let the distance of separation be $d$, increase be $x$ and dielectric length the be $t$.

When the dielectric is inserted:
$\frac{1}{C_{net}}=\frac{(d-t+x)}{\left(A{ϵ}_0\right)}+\frac{t}{\left(\kappa A{ϵ}_0\right)}$

And before the dielectric is inserted:

$\frac{1}{C_{initial}}=\frac{d}{\left(A{ϵ}_0\right)}$

Now, as per given condition:

$\frac{1}{C_{net}}=\frac{1}{C_{initial}}\Rightarrow d-t+x+\frac{t}{\kappa }=d\Rightarrow x=t(1-\frac{1}{\kappa })$

Therefore, increase in distance to restore original capacitance $=t(1-\frac{1}{\kappa })$$=4\times 0.8mm=3.2mm$