Question:

A cylindrical capacitor has charge \(Q\) and length \(L\). If both the charge and length of the capacitor are doubled, by keeping other parameters fixed, the energy stored in the capacitor

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If \(Q\) doubles and \(C\) doubles, energy \(U=\frac{Q^2}{2C}\) increases by factor 2.
Updated On: Jan 3, 2026
  • remains same
  • increases two times
  • decreases two times
  • increases four times
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The Correct Option is B

Solution and Explanation

Step 1: Recall energy stored in a capacitor.
\[ U = \frac{Q^2}{2C} \]
Step 2: Capacitance of cylindrical capacitor depends on length.
For a cylindrical capacitor:
\[ C \propto L \]
Step 3: Apply changes.
Given:
\[ Q' = 2Q,\quad L' = 2L \Rightarrow C' = 2C \]
Step 4: Compute new energy.
\[ U' = \frac{(2Q)^2}{2(2C)} = \frac{4Q^2}{4C} = \frac{Q^2}{C} \]
Original energy:
\[ U = \frac{Q^2}{2C} \]
Step 5: Compare ratio.
\[ \frac{U'}{U} = \frac{Q^2/C}{Q^2/(2C)} = 2 \]
So energy becomes double.
Final Answer:
\[ \boxed{\text{(B) increases two times}} \]
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