Derive an expression for the energy stored in parallel plate capacitor with air as the medium between its plates . Air is now re placed by dielectric medium of dielectric constants k. How does it charges the total energy of the capacitor if

a ) the capacitor remains connected to the same battery .

b ) the capacitor is disconnected from the battery.

When a capacitor is charged by a battery, it has to perform more and more work, so that it can deliver the same amount of charge over the plates as much is already present over it.
This work done is stored as electric energy inside the capacitor.
Now,
Work done in depositing dq amount of charge over the plates at potential difference V is given by

$$\begin{gathered} \mathrm{dW}=\mathrm{V} \mathrm{dq} \\ \mathrm{Since} \mathrm{q}=\mathrm{CV} \mathrm{or} \mathrm{V}=\mathrm{q}/\mathrm{C} \\ \mathrm{So}, \mathrm{dW}=\frac{\mathrm{q}}{\mathrm{C}}\mathrm{dq} \\ \mathrm{Total} \mathrm{work} \mathrm{done} \mathrm{in} \mathrm{depositing} \text{'}\mathrm{q}\text{'} \mathrm{charge} \mathrm{over} \mathrm{the} \mathrm{plates} \mathrm{is} \\ \mathrm{W}=\int \mathrm{dW} \\ ={\int }_0^{\mathrm{q}}\frac{\mathrm{q}}{\mathrm{C}}d\mathrm{q} \\ =\frac{1}{\mathrm{C}}{\int }_0^{\mathrm{q}}\mathrm{q}d\mathrm{q} \\ \mathrm{W}=\frac{1}{2}\frac{{\mathrm{q}}^2}{\mathrm{C}} \\ \mathrm{This} \mathrm{work} \mathrm{done} \mathrm{is} \mathrm{stored} \mathrm{as} \mathrm{energy} \mathrm{of} \mathrm{capacitor}. \\ \mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{CV}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{qV} \end{gathered}$$
(i) Energy of the capacitor increases on placing the dielectric slab between the plates of the capacitor when it remains connected to the battery.
(ii) Energy of the capacitor decreases on placing the dielectric slab between the plates of the capacitor when it is disconnected from the battery.