A Charged Capacitor of Capacitance 10 µF and p.d 30 v is Connected across an uncharged 50 µF capacitor. Calculate, i) The final p.d across the combination ii) The initial and final energies of the capacitors iii) Account for the difference in the initial & final energies.

The initial charge on $C_1$ is $Q_1 = C_1 V_1$.

The initial charge on $C_2$ is $Q_2 = C_2 V_2$. Since $C_2$ is uncharged, $V_2 = 0 \, V$.

When the capacitors are connected, they are in parallel. The total charge in the system is conserved.
Total initial charge $Q_{initial} = Q_1 + Q_2 = 300 \, \mu C + 0 \, \mu C = 300 \, \mu C$.
Total final charge $Q_{final} = Q_{initial} = 300 \, \mu C$.

For capacitors in parallel, the equivalent capacitance $C_{eq} = C_1 + C_2$.

The final potential difference across the combination is $V_f = \frac{Q_{final}}{C_{eq}}$.

The final p.d. across the combination is $\boxed{5 \, V}$.

The energy stored in a capacitor is $E = \frac{1}{2} C V^2$.
Initial energy in $C_1$:

Initial energy in $C_2$ is $0 \, J$ as it is uncharged.
Total initial energy:

After connection, both capacitors have the final potential difference $V_f = 5 \, V$.
Final energy in $C_1$:

Final energy in $C_2$:

Total final energy:

The initial energy is $\boxed{4.5 \times 10^{-3} \, J}$ and the final energy is $\boxed{0.75 \times 10^{-3} \, J}$.

When the charged capacitor is connected to the uncharged capacitor, charge flows from the charged capacitor to the uncharged capacitor until they reach a common potential. This flow of charge constitutes a current. Due to the resistance of the connecting wires, some of the initial stored electrical energy is dissipated as heat during this charge redistribution process.

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