The capacities of two capacitors are \( C_1 \) and \( C_2 \) and their respective potentials are \( V_1 \) and \( V_2 \). If they are connected with a wire, then the loss of energy is given by:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{C_1C_2\left(V_1 - V_2\right)^{2}}{2\left(C_1 + C_2\right)} \)
When the capacitors are joined by a wire, charge flows until the potentials are equal. The total charge is conserved, and the energy lost is the difference between the initial and final energies (it appears as heat in the wire).
Common potential
\( \displaystyle C_1V_1 + C_2V_2 = (C_1 + C_2)V \)
\( \displaystyle V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2} \)
Energies
\( \displaystyle U_i = \frac{1}{2}C_1V_1^{2} + \frac{1}{2}C_2V_2^{2},\qquad U_f = \frac{1}{2}(C_1 + C_2)V^{2} = \frac{(C_1V_1 + C_2V_2)^{2}}{2(C_1 + C_2)} \)
Loss
\( \displaystyle U_i - U_f = \frac{(C_1V_1^{2} + C_2V_2^{2})(C_1 + C_2) - (C_1V_1 + C_2V_2)^{2}}{2(C_1 + C_2)} \)
Expanding the numerator, the terms \( C_1^{2}V_1^{2} \) and \( C_2^{2}V_2^{2} \) cancel, leaving \( C_1C_2V_1^{2} + C_1C_2V_2^{2} - 2C_1C_2V_1V_2 \).
\( \displaystyle U_i - U_f = \frac{C_1C_2\left(V_1 - V_2\right)^{2}}{2\left(C_1 + C_2\right)} \)
Shortcut: Energy is proportional to \( V^{2} \) (or \( CV^{2} \)), so the loss must contain a square of a potential. Options (1), (2) and (4) are linear in the potentials and have the wrong dimension, so only Option (3) is possible. It is also zero when \( V_1 = V_2 \), as it should be, because then no charge flows.