STANDARDPhysics-Electrostatics
STANDARD Physics Capacitors MCQ Question
Type: MCQ-numerical-Medium
A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done?
A
6.25 × 10⁻⁶ J
B
3.75 × 10⁻⁶ J
C
2.16 × 10⁻⁶ J
D
2.55 × 10⁻⁶ J
Correct Answer
Option B
Detailed Explanation
The work done is calculated using the formula: Work done = U₀ - U = (q²/2) [1/C₀ - 1/C]. Substituting the given values, the work done is 3.75 × 10⁻⁶ J.
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