NEET 2022 · PhysicsStored energyPrevious Year Question
A capacitor of capacitance C = 900 pF is charged fully by 100 V battery B. Then B is disconnected and connected to another uncharged capacitor of capacitance C = 900 pF. The electrostatic energy stored by the system (b) is:
Answer: (C) 2.25 × 10⁻⁶ J. Answer: (c) Solution: U_i = ½CV² = ½(900 × 10⁻¹²)(100)² = 4.5 × 10⁻⁶ J.
- A.4.5 × 10⁻⁶ J
- B.3.25 × 10⁻⁶ J
- C.2.25 × 10⁻⁶ J✓
- D.1.5 × 10⁻⁶ J
Correct Answer
(C) 2.25 × 10⁻⁶ J
Solution & Explanation
Answer: (c) Solution: U_i = ½CV² = ½(900 × 10⁻¹²)(100)² = 4.5 × 10⁻⁶ J. Connecting an equal uncharged capacitor halves the energy → U_f = 2.25 × 10⁻⁶ J. → option (c).
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