A capacitor of capacitance $C = 900$ pF is charged fully by $100$ V battery $B$ as shown in figure (a). Then it is disconnected from the battery and connected to another uncharged capacitor of capacitance $C = 900$ pF as shown in figure (b). The electrostatic energy stored by the system (b) is
Answer: (D) $2.25 \times 10^{-6}$ J
Charge on the first capacitor: $Q = CV = 900 \times 10^{-12} \times 100 = 9 \times 10^{-8}$ C.
After connection the charge is shared between two equal capacitors, a total capacitance of $2C$, and the common voltage is $50$ V.
$$U = \frac{Q^2}{2(2C)} = \frac{(9 \times 10^{-8})^2}{4 \times 900 \times 10^{-12}} = \frac{81 \times 10^{-16}}{3.6 \times 10^{-9}} = 2.25 \times 10^{-6}\ \text{J}$$
(Half of the original $4.5 \times 10^{-6}$ J is lost as heat and radiation in the connecting wires.)
Solution by Sreeraj P, M.Sc Physics