A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery. Then the battery is disconnected, and an uncharged capacitor of capacitance C = 900 pF is connected in parallel. The electrostatic energy stored by the system is: [Diagram: Image]
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Question

A capacitor of capacitance C = 900 pF is charged fully by a 100 V battery. Then the battery is disconnected, and an uncharged capacitor of capacitance C = 900 pF is connected in parallel. The electrostatic energy stored by the system is: [Diagram: Image]

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G1 Physics · Accepted Answer

Step 1: Initial charge Q = CV = 900 pF × 100 V. Step 2: Equal uncharged capacitor connected in parallel gives combined 2C and final voltage V(final) = Q/(2C) = 50 V. Step 3: Total final stored energy U(final) = ½(2C)V(final)² = C(50)². Step 4: U(final) = 900×10⁻¹² × 2500 = 2.25 × 10⁻⁶ J.