A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy remaining in the system:
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A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy remaining in the system:

G1 Physics · Accepted Answer

Step 1: Initially charge CV and energy ½CV². Step 2: After identical capacitor is placed in parallel with no battery, charge remains CV and equivalent capacitance is 2C. Step 3: U(final) = Q²/(2×2C) = CV²/4 = ½U(initial). Thus energy is halved.