Question:medium

A capacitor of capacitance $8\mu\text{F}$ is fully charged by connecting it to a source of $200\text{ V}$. It is then disconnected from the supply and connected to an uncharged capacitor of capacitance $4\mu\text{F}$. The electrostatic energy lost in this sharing process is:

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Using the formula $\Delta U = \frac{1}{2}\frac{C_1C_2}{C_1+C_2}V^2$ allows you to solve the problem in a single line, avoiding the multi-step process of finding the total initial energy, calculating the common potential $V_c = \frac{C_1V_1}{C_1+C_2}$, and computing the final system energy.
Updated On: May 16, 2026
  • $5.33 \times 10^{-2}\text{ J}$
  • $21.34 \times 10^{-2}\text{ J}$
  • $10.67 \times 10^{-2}\text{ J}$
  • $3.53 \times 10^{-3}\text{ J}$
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The Correct Option is A

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