Energy of Charge Configurations

Chapter 1 — Electrostatics

If you want to assemble a set of charges from infinity, you have to do work against their mutual Coulomb forces. That work is stored as electrostatic potential energy, and it's one of two equivalent ways to count the total energy of a static configuration — the other being the energy density in the field.

Two charges

Bring \(q_1\) in from infinity: no work (nothing there yet). Bring \(q_2\) to distance \(r_{12}\) from \(q_1\): work against \(q_1\)'s field:

\[U_{12} = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r_{12}} = q_2 V_1(\vec r_2).\]

Positive if the charges have the same sign (repulsive — you fought the field), negative if opposite.

System of \(N\) charges

Assemble one at a time. Each new charge \(q_i\) costs work \(q_i V_{\text{already-there}}(\vec r_i)\). Summing and collecting:

\[U = \sum_{i

The factor of \(\tfrac12\) in the double-sum version accounts for double counting each pair. Equivalently,

\[U = \frac{1}{2}\sum_i q_i V(\vec r_i),\]

where \(V(\vec r_i)\) is the potential at \(q_i\)'s location due to all the other charges (not \(q_i\) itself).

Continuous limit

For a continuous distribution,

\[U = \frac{1}{2}\int \rho(\vec r)\, V(\vec r)\,dV.\]

No restriction to skip "self-energy" — for smooth distributions there are no divergences. For point charges, divergent self-energies appear and are discarded (they're absorbed into the particle's rest mass).

Self-energy of a uniform sphere

A classic result: the total energy required to assemble a uniform ball of radius \(R\) and total charge \(Q\) is

\[U = \frac{3}{5}\frac{kQ^2}{R}.\]

Derivation sketch: build the sphere shell by shell. When you've already built up to radius \(r\), adding a thin shell \(dq = \rho\cdot 4\pi r^2\,dr\) costs \(V(r)\,dq\) where \(V(r) = kq_{\text{so far}}/r = k\rho(\tfrac{4}{3}\pi r^3)/r\). Integrate \(dU = k(4\pi\rho/3)(4\pi\rho)r^4\,dr\) from 0 to \(R\) and simplify using \(Q = \tfrac43\pi R^3 \rho\).

Practice Problems