Question

A capacitor is charged by a battery and energy stored is $U$. Now the battery is removed and the distance between plates is increased to four times. The energy stored becomes

• A. $4U$
• B. $U$
• C. $3U$
• D. $2U$

Hint:

If charge remains constant, energy of capacitor is inversely proportional to capacitance.

Answer 1

The Correct Option is A

Step 1: Condition after removing battery.
When the battery is removed, charge on the capacitor remains constant.

Step 2: Relation between capacitance and plate separation.
Capacitance of a parallel plate capacitor is:
\[ C = \frac{\varepsilon_0 A}{d} \] If distance is increased to $4d$, new capacitance becomes:
\[ C' = \frac{C}{4} \]
Step 3: Energy stored in a capacitor.
\[ U = \frac{Q^2}{2C} \]
Step 4: New energy after increasing distance.
\[ U' = \frac{Q^2}{2C'} = \frac{Q^2}{2(C/4)} = 4U \]
Step 5: Conclusion.
Energy stored becomes $4U$.