Pre-Quantum Electrodynamics

Consider some distribution of charge which is held statically in place. How much work is needed to adiabatically move a small test charge \(q_t\) from one position to another, i.e. from point \({\bf a}\) to point \({\bf b}\)?

The work done in carrying this charge along some path is the negative of the electrical force in the direction of motion, so

\[ W = -\int_{{\bf a}}^{{\bf b}} {\bf F} \cdot d{\bf l} \]

Let's consider for simplicity a fixed source particle of charge \(q_s\) at position \({\bf r}_s \equiv {\bf 0}\). The work done against electrical forces when moving a unit charge test particle from \({\bf a}\) to \({\bf b}\) is then:

\[ -\int_{\bf a}^{\bf b} {\bf F} \cdot d{\bf l} = -\frac{q_t q_s}{4\pi \varepsilon_0} \int_{\bf a}^{\bf b} \frac{\bf r}{r^3} \cdot d{\bf r} \]

For the path, the 'angular' part is not contributing (see drawings in FLS II 4-3). The integral is thus purely radial,

\begin{equation} -\int_{\bf a}^{\bf b} {\bf F} \cdot d{\bf l} = -\frac{q_t q_s}{4\pi \varepsilon_0} \int_{r_a}^{r_b} \frac{dr}{r^2} = \frac{q_t q_s}{4\pi \varepsilon_0} \left(\frac{1}{r_b} - \frac{1}{r_a}\right). \tag{Wab}\label{Wab} \end{equation}

As should be clear by now, this result does not depend on the path (if it did, we'd have a perpetuum mobile when going from \({\bf r}_a\) to \({\bf r}_b\) one way, and coming back another).

When thinking about the energy of this pair of charges, we think of starting from an initial configuration where the charges are infinitely distant (which we associate to zero energy), and thus set \({\bf r}_a = \infty\). For a pair of charges, the energy is thus

\[ W = \frac{1}{4\pi \varepsilon_0} \frac{q_t q_s}{|{\bf r}_t - {\bf r}_s|} \]

By the superposition principle, the energy of a generic assembly of charges \(\{ q_i \}\), \(i = 1, .. n\) (sitting at positions \({\bf r}_i\)) is obtained by the pairwise sum (counting each pair only once) of pairwise energies:

\[ W = \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n \sum_{j > i}^n \frac{q_i q_j}{|{\bf r}_i - {\bf r}_j|} = \frac{1}{8\pi \varepsilon_0} \sum_{i=1}^n \sum_{j \neq i}^n \frac{q_i q_j}{|{\bf r}_i - {\bf r}_j|} \label{Gr(2.41)} \]

where in the second equality we have symmetrized the sums for convenience.

• PM 1.6
• FLS II 8-4