1. Concept: When a charged capacitor is disconnected, its total charge \( Q \) becomes trapped and strictly conserved. Connecting an identical uncharged capacitor in parallel forces the charge to distribute evenly.
2. Formula: $$ U = \frac{Q^2}{2C} $$
3. Solution: - Initial state: Single capacitor has charge \( Q \). Energy is \( U = \frac{Q^2}{2C} \).
- Second state: Because they are identical and parallel, the trapped total charge \( Q \) splits perfectly. The new charge on EACH individual capacitor is \( Q' = Q/2 \).
- Calculate the new energy for ONE capacitor: \( U_{each} = \frac{(Q/2)^2}{2C} = \frac{Q^2/4}{2C} = \frac{1}{4} \left( \frac{Q^2}{2C} \right) \).
- Since the original \( U = \frac{Q^2}{2C} \), the final energy in each is exactly \( U/4 \).
4. Why other options are incorrect: Option A (U/2) is the total combined energy left in the entire system, representing the 50% energy lost to heat during charge redistribution. The question asks for the energy in
each capacitor, which is half of the remaining total.
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