Question:
A capacitor of unknown capacitance is connected across a battery of $V$ volt. The charge stored in it is $Q$ coulomb. When potential across the capacitor is reduced by $V'$ volt, the charge stored in it becomes $Q'$ coulomb. The potential $V$ is
A capacitor of unknown capacitance is connected across a battery of $V$ volt. The charge stored in it is $Q$ coulomb. When potential across the capacitor is reduced by $V'$ volt, the charge stored in it becomes $Q'$ coulomb. The potential $V$ is
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Always express unknown capacitance using $Q = CV$ to eliminate it from equations.
Updated On: Feb 4, 2026
- $\dfrac{QV'}{(Q+Q')}$
- $\dfrac{(Q+Q')}{QV'}$
- $\dfrac{(Q-Q')}{QV'}$
- $\dfrac{QV'}{(Q-Q')}$
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The Correct Option is D
Solution and Explanation
Step 1: Use basic relation of capacitance.
For a capacitor: \[ Q = CV \] Step 2: Write equations for both situations.
Initially: \[ Q = CV \] After reducing potential by $V'$: \[ Q' = C(V - V') \] Step 3: Subtract the two equations.
\[ Q - Q' = CV' \] Step 4: Solve for $V$.
From $Q = CV \Rightarrow C = \dfrac{Q}{V}$
Substitute into previous equation: \[ Q - Q' = \dfrac{Q}{V}V' \] \[ V = \dfrac{QV'}{(Q - Q')} \]
For a capacitor: \[ Q = CV \] Step 2: Write equations for both situations.
Initially: \[ Q = CV \] After reducing potential by $V'$: \[ Q' = C(V - V') \] Step 3: Subtract the two equations.
\[ Q - Q' = CV' \] Step 4: Solve for $V$.
From $Q = CV \Rightarrow C = \dfrac{Q}{V}$
Substitute into previous equation: \[ Q - Q' = \dfrac{Q}{V}V' \] \[ V = \dfrac{QV'}{(Q - Q')} \]
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