A capacitor of unknown capacitance $\mathrm{C}$ is connected across a battery of $\mathrm{V}$ volt. The…

A capacitor of unknown capacitance $\mathrm{C}$ is connected across a battery of $\mathrm{V}$ volt. The charge stored in it becomes $Q$ coulomb. When potential across the capacitor is reduced by $\mathrm{V}^{\prime}$ volt, the charge stored in it becomes Q'coulomb. The capacitance C is
  1. $\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\sqrt{\mathrm{V}^{\prime}}}$
  2. $\frac{\mathrm{V}^{\prime}}{\mathrm{Q}-\mathrm{Q}^{\prime}}$
  3. $\frac{\mathrm{Q}+\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}$
  4. $\frac{\mathrm{Q}-\mathrm{Q}^{\prime}}{\mathrm{V}^{\prime}}$

Solution

$\mathrm{Q}=\mathrm{CV}$ $\mathrm{Q}^{\prime}=\mathrm{C}\left(\mathrm{V}-\mathrm{V}^{\prime}\right)=\mathrm{CV}=\mathrm{CV}^{\prime}$ $\mathrm{Q}^{\prime}=\mathrm{Q}-\mathrm{CV}^{\prime}$ $\therefore \mathrm{CV}^{\prime}=\mathrm{Q}-\mathrm{Q}^{\prime}$ $\mathrm{C}=\frac{Q-Q^{\prime}}{V^{\prime}}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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