Physics - Capacitor Question with Solution | TestHub

When two charged conductors are connected, charge flows until their potentials become equal. This redistribution of charge generally involves a loss of electrostatic potential energy, which is dissipated as heat or electromagnetic radiation. An exception occurs if their initial potentials are already equal, meaning $Q_1/R_1=Q_2/R_2$, in which case no charge flows and no energy is lost.

The initial potential energy of the system is:

$$U_i=\frac{1}{2}k\frac{Q_1^2}{R_1}+\frac{1}{2}k\frac{Q_2^2}{R_2}$$

After connection, the total charge $Q=Q_1+Q_2$ is redistributed such that the final potentials are equal: $V_f=k{Q}_1^{\prime }/R_1=k{Q}_2^{\prime }/R_2$.

The final potential energy is:

$$U_f=\frac{1}{2}k\frac{(Q_1^{\prime }{)}^2}{R_1}+\frac{1}{2}k\frac{(Q_2^{\prime }{)}^2}{R_2}$$

It can be shown that $U_f\le U_i$, with equality only if the initial potentials were already equal ($Q_1/R_1=Q_2/R_2$). Therefore, there is always a decrease in the energy of the system unless $Q_1{R}_2=Q_2{R}_1$.