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In a parallel-plate capacitor, the distance between the plates is d and the potential difference across the plates is V. The energy stored per unit volume between the plates is

1. \(\frac{Q^2}{2V}\)
2. \(\frac{\varepsilon_0 V^2}{2d}\)
3. \(\frac{\varepsilon_0 V^2}{2d^2}\)
4. \(\frac{\varepsilon_0 V^2}{d}\)

Correct answer: \(\frac{\varepsilon_0 V^2}{2d^2}\)

Solution

The energy density in an electric field is \(u = \frac{1}{2}\varepsilon_0 E^2\). For a parallel-plate capacitor, the field between the plates is \(E = V/d\), so \(u = \frac{1}{2}\varepsilon_0 (V/d)^2 = \frac{\varepsilon_0 V^2}{2d^2}\).

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