A capacitor of capacitance C = 900 pF is charged fully by 100 V battery. Then the battery is disconnected and the capacitor is connected to another un - Physics MCQ NEET 2022 2026 | ExamTest.live

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A capacitor of capacitance C = 900 pF is charged fully by 100 V battery. Then the battery is disconnected and the capacitor is connected to another uncharged capacitor of capacitance C = 900 pF . The electrostatic energy stored by the system is:

Solution & Step-by-step Explanation

Initial energy U_{i} = \frac{1}{2} C V^{2} : U_{i} = \frac{1}{2} \times (900 \times 1 0^{- 12}) \times (100)^{2} = 4.5 \times 1 0^{- 6} J When connected to an identical uncharged capacitor, the common potential becomes V^{'} = V /2 = 50 V .Total equivalent capacitance C_{e q} = C + C = 2 C = 1800 pF .Final energy U_{f} = \frac{1}{2} C_{e q} (V^{'})^{2} : U_{f} = \frac{1}{2} \times (1800 \times 1 0^{- 12}) \times (50)^{2} = 900 \times 1 0^{- 12} \times 2500 = 2250000 \times 1 0^{- 12} U_{f} = 2.25 \times 1 0^{- 6} J

Try it yourself before checking the explanation above.

A capacitor of capacitance C = 900 pF is charged fully by 100 V battery. Then the battery is disconnected and the capacitor is connected to another uncharged capacitor of capacitance C = 900 pF . The electrostatic energy stored by the system is:

4.5 \times 1 0^{- 6} J

3.25 \times 1 0^{- 6} J

2.25 \times 1 0^{- 6} J

1.5 \times 1 0^{- 6} J

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