Share a capacitor’s charge and track its energy

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Chemical and Physical Foundations

A capacitor of capacitance $C$ is charged to a potential difference $V$ by a battery. The capacitor is disconnected from the battery and then connected in parallel to an identical, initially uncharged capacitor. What fraction of the initial energy stored in the charged capacitor is stored in the two-capacitor system after electrostatic equilibrium is reached?

Select one answer.

\frac{1}{4}

\frac{1}{2}

1

2

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Answer: B ( $\frac{1}{2}$ )

Initially, the capacitor has charge $Q=CV$ and stores energy $U_i=\frac{1}{2}CV^{2}$ . After the battery is disconnected, the total charge is conserved. Connecting the identical capacitors in parallel produces an equivalent capacitance of $2C$ , so the final potential difference is $V_f$ ${}=\frac{Q}{2C}$ ${}=\frac{V}{2}$ . The final energy is therefore $U_f$ ${}=\frac{1}{2}(2C)\left(\frac{V}{2}\right)^{2}$ ${}=\frac{1}{4}CV^{2}$ ${}=\frac{1}{2}U_i$ . Thus, the two-capacitor system stores $\frac{1}{2}$ of the initial energy.

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Chemical and Physical Foundations

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