A capacitor of capacitance $$50$$ pF is charged by $$100$$ V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is ______ nJ.

A capacitor of 50 pF is charged by a 100 V source, then connected to another uncharged identical capacitor.

The initial energy stored in the capacitor is $$U_i = \frac{1}{2}CV^2 = \frac{1}{2} \times 50 \times 10^{-12} \times (100)^2$$. Substituting the values gives $$U_i = \frac{1}{2} \times 50 \times 10^{-12} \times 10000 = 250 \times 10^{-9} \text{ J} = 250 \text{ nJ}$$.

When this capacitor is connected to an identical uncharged one, the total charge is shared equally and the final voltage across each capacitor is $$V_f = \frac{Q_{total}}{C_{total}} = \frac{CV}{2C} = \frac{V}{2} = 50 \text{ V}$$.

The total energy stored in the two capacitors after connection is $$U_f = \frac{1}{2}(2C)V_f^2 = \frac{1}{2} \times 2 \times 50 \times 10^{-12} \times (50)^2$$, which gives $$U_f = 50 \times 10^{-12} \times 2500 = 125 \times 10^{-9} \text{ J} = 125 \text{ nJ}$$.

The energy loss is the difference between the initial and final energies: $$\Delta U = U_i - U_f = 250 - 125 = 125 \text{ nJ}$$.

The electrostatic energy loss is 125 nJ.