Energy Stored in a Capacitor

When you charge a capacitor, you do work against the rising voltage. That work is stored as electric potential energy — not in the plates, but in the field between them.

Concept

A capacitor charged to voltage $V$ with charge $Q$ stores energy $U = \frac{1}{2}QV = \frac{1}{2}CV^2 = \frac{Q^2}{2C}.$ The three forms are equivalent (use $Q = CV$).

Energy density of the electric field: $u = \frac{1}{2}\varepsilon_0 E^2.$ With dielectric, $\varepsilon_0 \to K\varepsilon_0$.

This is a local quantity — every cubic metre of space with a field $E$ holds this much energy, independent of the source. The energy in any capacitor equals the integral of $u$ over the space where field is present.

Derivation

$U = \tfrac{1}{2}CV^2$: during charging, suppose at some instant the charge is $q$ and voltage $v = q/C$. To move an additional $dq$ from one plate to the other against this voltage requires work $dW = v\,dq = \frac{q}{C}dq.$ Integrate from $0$ to $Q$: $U = \int_0^Q \frac{q}{C}dq = \frac{Q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}QV.$ The factor of $\tfrac{1}{2}$ is essential — naively writing $U = QV$ overcounts.

Energy density. For a parallel-plate capacitor with field $E$, plate area $A$, gap $d$: $U = \frac{1}{2}CV^2 = \frac{1}{2}\frac{\varepsilon_0 A}{d}(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 (Ad).$ $Ad$ is the volume between plates, so $u = U/(\text{volume}) = \tfrac{1}{2}\varepsilon_0 E^2.$

Worked Example

A $50\,\mu F$ capacitor charged to $100\,V$: $U = \tfrac{1}{2}(50\times 10^{-6})(100)^2 = 0.25\,J.$

Same capacitor: charge $Q = CV = 5000\,\mu C = 5\,mC$. Check: $U = Q^2/(2C) = (5\times 10^{-3})^2/(2\times 50\times 10^{-6}) = 0.25\,J$. ✓

Energy density. A field of $E = 10^5\,V/m$ in vacuum: $u = \tfrac{1}{2}(8.854\times 10^{-12})(10^5)^2 \approx 0.044\,J/m^3.$

Two capacitors connected together. A $4\,\mu F$ at $100\,V$ and an uncharged $4\,\mu F$ are joined in parallel. Initial energy: $\tfrac{1}{2}(4\,\mu F)(100)^2 = 0.02\,J$. Final voltage equalizes: $Q_\text{total} = 400\,\mu C$ shared equally over total $8\,\mu F$, giving $V = 50\,V$. Final energy: $\tfrac{1}{2}(8\,\mu F)(50)^2 = 0.01\,J$. Half the energy is lost as heat/radiation in the wires.

Common Confusions

• $U = \tfrac{1}{2}QV$, not $QV$. The factor of $\tfrac{1}{2}$ comes from the integral over charging.
• Energy is stored in the field, not on the plates. Hence energy density.
• Energy lost when capacitors connect. It is real heat dissipation — even ideal wires lose half the energy via radiation if resistance vanishes.
• Inserting a dielectric with battery connected: energy increases ($U = \tfrac{1}{2}CV^2$, $C$ rises, $V$ fixed). The battery supplies the extra energy.
• Inserting a dielectric with battery disconnected: energy decreases ($U = Q^2/2C$, $C$ rises, $Q$ fixed). Where does it go? Into the dielectric being sucked in — mechanical work.

Key Takeaways

• $U = \tfrac{1}{2}CV^2 = QV/2 = Q^2/(2C)$.
• Energy density: $u = \tfrac{1}{2}\varepsilon_0 E^2$ (vacuum); $\tfrac{1}{2}K\varepsilon_0 E^2$ in dielectric.
• Field stores energy throughout space, not on plates.
• Joining capacitors at different voltages dissipates energy.