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The parallel-plate capacitor shown in the figure has movable plates. The capacitor is charged so that the energy stored in it is E when the plate separation is d. The capacitor is then isolated electrically and the plates are moved such that the plate separation becomes 2d.
At this new plate separation, what is the energy stored in the capacitor, neglecting fringing effects?
- 2E
- √2E
- E
- E/2
Correct answer: 2E
Solution
After isolation the charge Q is fixed. C=eps*A/d halves when separation goes d->2d. Energy U=Q^2/(2C) therefore doubles to 2E (work done pulling the attracting plates apart). The stored sqrt(2)E is wrong; the answer is 2E.
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