Charge Redistribution and Sharing

When two charged capacitors are connected — charge flows, energy is lost.

Connecting Two Capacitors

When two charged capacitors are connected to each other (positive plate to positive plate, negative to negative), charge redistributes until the potential difference across both is equal.

Let capacitor 1 have capacitance \( C_1 \) initially charged to \( V_1 \) (charge \( Q_1 = C_1 V_1 \)), and capacitor 2 have \( C_2 \) initially charged to \( V_2 \) (charge \( Q_2 = C_2 V_2 \)).

Common Voltage After Sharing

By conservation of charge (charge cannot be created or destroyed, only redistributed):

\[ Q_1 + Q_2 = Q_1' + Q_2' = (C_1 + C_2) V_f \] \[ \boxed{V_f = \frac{Q_1 + Q_2}{C_1 + C_2} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}} \]

This is a weighted average of the initial voltages, weighted by capacitance.

Final Charges on Each Capacitor

After equilibrium, both capacitors are at \( V_f \):

\[ Q_1' = C_1 V_f = \frac{C_1(C_1 V_1 + C_2 V_2)}{C_1 + C_2} \] \[ Q_2' = C_2 V_f = \frac{C_2(C_1 V_1 + C_2 V_2)}{C_1 + C_2} \]

Special case — one capacitor initially uncharged (\( V_2 = 0 \)):

\[ V_f = \frac{C_1 V_1}{C_1 + C_2}, \quad Q_1' = \frac{C_1^2 V_1}{C_1 + C_2}, \quad Q_2' = \frac{C_1 C_2 V_1}{C_1 + C_2} \]

Energy Lost During Charge Sharing

The total initial energy is:

\[ U_i = \frac{Q_1^2}{2C_1} + \frac{Q_2^2}{2C_2} = \frac{1}{2}C_1 V_1^2 + \frac{1}{2}C_2 V_2^2 \]

The total final energy is:

\[ U_f = \frac{1}{2}(C_1 + C_2)V_f^2 = \frac{(C_1 V_1 + C_2 V_2)^2}{2(C_1 + C_2)} \]

The energy lost is:

\[ \Delta U = U_i - U_f = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)} \]

Energy is always lost when two capacitors at different potentials are connected — even through ideal, resistance-free wires. The energy is dissipated as electromagnetic radiation (or heat in any real resistance). This loss is proportional to \( (V_1 - V_2)^2 \) and is zero only when the initial voltages are equal.

Opposite Polarity Connection

If the capacitors are connected with opposite polarities (positive plate of one to negative plate of the other):

\[ Q_{\text{net}} = Q_1 - Q_2 = C_1 V_1 - C_2 V_2 \] \[ V_f = \frac{C_1 V_1 - C_2 V_2}{C_1 + C_2} \]

The sign of \( V_f \) indicates which capacitor's polarity "wins." If \( C_1 V_1 = C_2 V_2 \), the final voltage is zero — complete cancellation of charge.

JEE Key Points

Use conservation of charge to find \( V_f \) — not conservation of energy (energy is not conserved due to radiation losses).
The formula \( \Delta U = C_1 C_2 (V_1 - V_2)^2 / [2(C_1 + C_2)] \) gives the energy dissipated — always positive.
For same-polarity connection: charges add. For opposite-polarity: charges subtract.
If \( C_1 = C_2 = C \) and \( V_2 = 0 \): \( V_f = V_1/2 \), energy lost = \( CV_1^2/4 = U_i/2 \). Half the energy is always lost.