A capacitor of capacitance C is charged to a potential difference… - Vidyadip

Question

A capacitor of capacitance $C$ is charged to a potential difference $V_0$. The charging battery is removed and the capacitor is now connected to an uncharged capacitor of unknown capacity, the potential difference across the combination becomes $V$. The unknown capacitance is

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Answer

The capacitor $\mathrm{C}$ after charging to a potential difference $\mathrm{V}_{\mathrm{0}}$ is disconnected from the battery. It is now connected with another capacitor of capacitance $\mathrm{C}^{\prime}(\text { say }) .$ The charge of first capacitor distributes over the second one such that both the capacitor acquire same potential difference $\mathrm{V}(\text{say}).$ This common potential difference $\mathrm{V}$ is given as

$\mathrm{V}=\frac{\mathrm{C}_{1} \mathrm{V}_{1}+\mathrm{C}_{2} \mathrm{V}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}}$

$\mathrm{V}=\frac{\mathrm{CV}_{0}+\mathrm{C}_{2} \times 0}{\mathrm{C}+\mathrm{C}_{2}}$

$\mathrm{C}+\mathrm{C}_{2}=\frac{\mathrm{CV}_{0}}{\mathrm{V}} \Rightarrow \mathrm{C}_{2}=\mathrm{C}\left(\frac{\mathrm{V}_{0}}{\mathrm{V}}-1\right)$

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