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A charged capacitor stores energy in its electric field. Three equivalent formulas: U = \frac{1}{2}$$CV^2 = $\frac{1}{2}$QV = Q^$\frac{2}{2C}$. The factor 1/2 arises because the average voltage during charging is V/2.

Energy density (per unit volume) in an electric field: u = $\frac{1}{2}$$epsilon_0$$E^2$ (vacuum) or u = $\frac{1}{2}$K$epsilon_0$$E^2$ (dielectric). Total energy = integral of u over the volume. For parallel plates: U = u * (Ad) = $\frac{1}{2}$$epsilon_0$$\frac{V}{d}$^2*Ad = \frac{1}{2}$$CV^2.

When charging from a constant voltage source (battery), the battery delivers energy QV = $CV^2$, but only \frac{1}{2}$$CV^2 is stored. The other half is dissipated as heat in the circuit (regardless of resistance). This 50% charging efficiency is a fundamental result.

Force between plates: F = Q^$\frac{2}{2*epsilon_0*A}$ = $\frac{1}{2}$$epsilon_0$$E^2$A (attractive). Electrostatic pressure: P = sigma^$\frac{2}{2*epsilon_0}$ = $\frac{1}{2}$$epsilon_0$$E^2$. Each plate is in the field of the other plate ($E_{other}$ = sigma/2$epsilon_0$), not the total field.

Self-energy of charge distributions: solid sphere U = 3kQ^$\frac{2}{5R}$; shell U = kQ^$\frac{2}{2R}$. These represent the energy needed to assemble the distribution from infinitesimal charge elements. As R -> 0, self-energy diverges — the classical self-energy problem.

JEE frequently combines energy concepts with dielectric insertion, plate separation changes, or charge redistribution between capacitors.