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.

Found an issue with this question?

Explore More Questions