Question

A potential difference is applied to the plate of a capacitor filled with an insulator, with energy stored in the capacitor as U. The capacitor is disconnected and the insulator is pulled out now. The work done in pulling out the insulator against electric field is 4U. Find the dielectric constant of the insulator. ___

Answer 1

Let the initial capacitance be $C_1$ =$\frac{k{\epsilon }_{0}A}{d}$

Now we can write,

$V^{}=\frac{1}{2}{C}_1{V_1}^{}$

Now let the initial charge be Q and the voltage be $V_1$

Now when the battery is removed charge Q will become constant.

When the insulation is removed capacitance decreases by k

Now, $C_2=\frac{{\epsilon }_{0}A}{d}{V}_2=V_{1}k$

Now $V=\frac{1}{2}{C}_1{v_1}^2$

Final dielectric constant = $\frac{1}{2}\frac{C-1}{k}{\left(V_{1}k\right)}^2=k\mu$

It is given in the question that the work done is $4\mu$

Now,

$4\mu =k\mu -\mu$

After solving

k = 5

Hence the dielectric constant of the insulator is 5.