A parallel plate capacitor of capacitance 200 µF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be ......... J.

Given, capacity of capacitor, C = 200 µF

EMF of battery = Potential difference across capacitor, V = 200 V

Dielectric constant of slab, K = 2

        ${\mathrm{U}}_1=\frac{1}{2}{\mathrm{CV}}^2=\frac{1}{2}\times 200\times (200{)}^2\mathrm{\mu J}=4\mathrm{J}$

New capacity with dielectric slab,

        $\mathrm{C}\text{'}=\mathrm{KC}=200\times 2=400 \mathrm{\mu F}$

Since, the battery remains connected.

Then,           V′ =V = 200 V

Now, electrostatic energy in capacitor with dielectric

         $\begin{array}{l}{\mathrm{U}}_2=\frac{1}{2}\mathrm{C}\text{'}\mathrm{V}{\text{'}}^2=\frac{1}{2}\times 400\times (200{)}^2 \mathrm{\mu J}\\ =8 \mathrm{J}\end{array}$

Hence, change in the electrostatic energy,

          $\mathrm{\Delta U}={\mathrm{U}}_2-{\mathrm{U}}_1=8-4=4 \mathrm{J}$

Thus, the correct answer is 4.