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Formulas/physics/Electrostatic Potential/Energy Stored in Capacitor (3 Forms)

Energy Stored in Capacitor (3 Forms)

Energy stored in a charged capacitor. Three equivalent forms depending on which pair of (Q, C, V) is known. The factor ½ arises because V builds up gradually as charge is added.
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Derivation

Derivation by integration

When charge qq is already on the capacitor, the potential is v=q/Cv = q/C. The work done adding an infinitesimal charge dqdq against this potential:

dW=v dq=qC dqdW = v\, dq = \frac{q}{C}\, dq

Total work done charging from 00 to QQ:

U=∫0QqC dq=1C⋅Q22U = \int_0^Q \frac{q}{C}\, dq = \frac{1}{C}\cdot\frac{Q^2}{2} U=Q22C=12CV2=QV2\boxed{U = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{QV}{2}}

The three forms follow from substituting Q=CVQ = CV:

Q22C=(CV)22C=CV22,QV2=CV⋅V2=CV22✓\frac{Q^2}{2C} = \frac{(CV)^2}{2C} = \frac{CV^2}{2}, \qquad \frac{QV}{2} = \frac{CV \cdot V}{2} = \frac{CV^2}{2} \checkmark

The factor ½

The factor 12\frac{1}{2} arises because the potential builds up from 0 to VV as charge accumulates. The average potential during charging is V/2V/2. This is why U=12QVU = \frac{1}{2}QV, not QVQV.

Where is the energy stored?

The energy is stored in the electric field between the plates — not in the charges themselves. This becomes explicit in the energy density formula (see ep20-energy-density).

Remember
When a charged capacitor is connected to another uncharged one, energy is lost to heat even if there is no resistance. This is because current flows during charge redistribution.