Energy in a Capacitor
Chapter 3 — Conductors
Charging a capacitor requires work — you're pushing like charges together. That work is stored as electrostatic energy, retrievable by discharging.
Derivation via charging
Imagine charging the capacitor from \(q = 0\) to final charge \(Q\), one infinitesimal piece at a time. When the charge is currently \(q\), the voltage is \(\Delta V(q) = q/C\). Moving another piece \(dq\) across that voltage costs
\[dU = \Delta V(q)\,dq = \frac{q}{C}\,dq.\]
Integrating from 0 to \(Q\):
\[U = \int_0^Q \frac{q}{C}\,dq = \frac{Q^2}{2C}.\]
The three equivalent expressions
Using \(Q = C\,\Delta V\), you can rewrite:
\[\boxed{\ U = \frac{Q^2}{2C} = \tfrac{1}{2}C\,\Delta V^2 = \tfrac{1}{2}Q\,\Delta V.\ }\]
They're all the same energy, expressed in different variables. Pick the one that matches what's held fixed in your problem.
Energy as field energy
The energy is also \(U = \int \tfrac{1}{2}\epsilon_0 E^2\,dV\) — integrated over all space. For a parallel-plate capacitor the field is uniform between the plates and zero outside, so
\[U = \tfrac{1}{2}\epsilon_0 E^2\cdot (Ad) = \tfrac{1}{2}\epsilon_0\left(\frac{\Delta V}{d}\right)^2 A d = \tfrac{1}{2}\,\frac{\epsilon_0 A}{d}\,\Delta V^2 = \tfrac{1}{2}C\,\Delta V^2. ✓\]
The two viewpoints (\(Q^2/2C\) vs integrated energy density) give the same answer — as they must.
Force from energy (fixed Q)
If the capacitor is disconnected (\(Q\) fixed), the force between the conductors along a coordinate \(x\) is
\[F_x = -\frac{\partial U}{\partial x}\bigg|_Q = -\frac{\partial}{\partial x}\left(\frac{Q^2}{2C(x)}\right) = \frac{Q^2}{2C^2}\frac{\partial C}{\partial x}.\]
Force from energy (fixed V)
If the capacitor is connected to a battery (\(\Delta V\) fixed), the battery does work as charge flows, and the correct formula is
\[F_x = +\frac{\partial U}{\partial x}\bigg|_{\Delta V} = +\tfrac{1}{2}\Delta V^2\,\frac{\partial C}{\partial x}.\]
The sign flip between the two cases is easy to get wrong on an exam. It's not that the forces are opposite — it's that the work-energy accounting is different (the battery is a reservoir). In fact, plugging \(Q = C\Delta V\) into the two expressions gives the same force, as they must. The sign difference in the formula accounts for battery work.
Practice Problems
Hint
Solution
\(U = \tfrac{1}{2}(10\times 10^{-6})(12)^2 = 7.2\times 10^{-4}\,\mathrm{J}\).
Answer: \(0.72\,\mathrm{mJ}\).
Hint
Solution
Initial: \(Q = C_0 V_0\), \(U_0 = \tfrac{1}{2}C_0 V_0^2\). After: \(C = C_0/2\), \(Q\) unchanged, \(U = Q^2/(2C) = (C_0 V_0)^2/(C_0) = C_0 V_0^2 = 2U_0\).
You did external work pulling the plates apart against their attraction; that work shows up as extra stored energy.
Answer: \(U = 2U_0\); the factor of 2 came from the mechanical work you did.
Hint
Solution
\(\Delta C \approx (\epsilon_0 A/d^2)\delta\) (capacitance rises). Extra charge: \(\Delta Q = V\Delta C\). Battery work: \(W_\text{bat} = V\Delta Q = V^2\Delta C\). Change in stored energy: \(\Delta U = \tfrac{1}{2}V^2\Delta C\). Difference: \(W_\text{bat} - \Delta U = \tfrac{1}{2}V^2\Delta C\) — this is the work done on the plates by the attractive force.
Answer: Battery supplies \(V^2\Delta C\); half goes to stored field energy, half to mechanical work on the plates.
Hint
Solution
\[U = \int_{R_1}^{R_2} \tfrac{1}{2}\epsilon_0 \left(\frac{kQ}{r^2}\right)^2 4\pi r^2\,dr = \tfrac{1}{2}\epsilon_0\cdot 4\pi k^2 Q^2 \int_{R_1}^{R_2}\frac{dr}{r^2} = \frac{Q^2}{8\pi\epsilon_0}\left(\frac{1}{R_1} - \frac{1}{R_2}\right).\]
Compare: \(C = 4\pi\epsilon_0 R_1 R_2/(R_2 - R_1)\), so \(\frac{Q^2}{2C} = \frac{Q^2(R_2-R_1)}{8\pi\epsilon_0 R_1 R_2} = \frac{Q^2}{8\pi\epsilon_0}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\). ✓
Answer: \(U = \frac{Q^2}{8\pi\epsilon_0}\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\); agrees with \(Q^2/(2C)\).