A parallel-plate capacitor of capacitance $40\ \mu\text{F}$ is connected to a $100\ \text{V}$ power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant $K = 2$. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are
Answer: (A) $4\ \text{mC}$ and $0.2\ \text{J}$
The supply stays connected, so $V = 100\ \text{V}$ is fixed and $C$ becomes $KC = 80\ \mu\text{F}$.
Charge: $Q_1 = 40\times10^{-6}\times100 = 4\ \text{mC}$, $Q_2 = 80\times10^{-6}\times100 = 8\ \text{mC}$. Extra charge $= 4\ \text{mC}$.
Energy: $U_1 = \tfrac{1}{2}(40\times10^{-6})(100)^2 = 0.2\ \text{J}$, $U_2 = \tfrac{1}{2}(80\times10^{-6})(100)^2 = 0.4\ \text{J}$. Change $= 0.2\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics