A capacitor is filled with an insulator and a potential difference is applied to its plates. The energy stored in the capacitor is U. The capacitor is disconnected from the power source and the insulator is pulled out of it. The work done against the electric force in pulling out the insulator in this case is 4U. What is the dielectric constant of the insulator?

5

The charge remains constant in both the cases.

The capacitor originally had the energy U and the work required to pull the insulator out is 4U so the final energy is equal to 5U.

$U_f = U_i + W = U + 4U = 5U$

The CAPACITANCE OF AN INSULATOR WITH DIELECTRIC IS $C = \frac{k\epsilon_0 A}{d}$

$ \text{The Energy stored in the capacitor is } U_i= \frac{q^2}{2C}$

$ C_0 = \frac{\epsilon_0 A}{d}$

Energy Stored will be $ U_f = \frac{q^2}{2C_0} = 5U$

$ \Rightarrow \frac{q^2}{2C_0} = 5 \frac{q^2}{2C}$

$ C = 5C_0$

$ \Rightarrow k = 5$