Question:

The energy stored in a capacitor is $W$. To double the charge on the plates of the capacitor, the additional work to be done is

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Energy stored in a capacitor is proportional to the square of voltage. Doubling the charge increases the energy fourfold.
Updated On: Jun 4, 2025
  • $W$
  • $4W$
  • $\dfrac{4}{3}W$
  • $3W$
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The Correct Option is D

Solution and Explanation

Step 1: Initial energy stored
\[ U_1 = \frac{1}{2}CV^2 = W \] Step 2: Final energy when charge is doubled
Since \(Q = CV\), then \(V = \frac{Q}{C}\). If charge is doubled (\(Q \rightarrow 2Q\)), then voltage doubles too: \(V \rightarrow 2V\). So the new energy:
\[ U_2 = \frac{1}{2}C(2V)^2 = \frac{1}{2}C \cdot 4V^2 = 4 \cdot \frac{1}{2}CV^2 = 4W \] Step 3: Additional work done
\[ \text{Additional work} = U_2 - U_1 = 4W - W = 3W \]
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