Question

A capacitor is filled with an insulator and a certain potential difference is applied to its plates. The energy stored in the capacitor is $U$. Now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. The work performed against the forces of the electric field in pulling out the insulator is $4U$. Then the dielectric constant of the insulator is :

• A. $4$
• B. $8$
• C. $5$
• D. $3$

Answer 1

The correct option is C $5$

Let the initial capacitance be $C_1=\frac{k{\epsilon }_{0}A}{d}$
$\Rightarrow V_1=\frac{1}{2}{C}_1{V}_1^2$
Let initial charge be Q and voltage be $V_1$
Now when the battery is removed Q becomes constant. When the insulation is removed c, decreases by k $\therefore$ V increases by k.
$C_2=\frac{{\epsilon }_{0}A}{d}{V}_2=V_{1}k.$
Now $V=\frac{1}{2}{C}_1{v}_1^2$
final ${\mu }_2=\frac{1}{2}\frac{C-1}{k}(V_{1}k{)}^2=k\mu$
given work done $=4\mu$
$\Rightarrow 4\mu =k\mu -\mu$
$\Rightarrow k=5$