25.2.3 Multiple Point Charges

If more than two charges are present, their potential energy is the sum of the potential energies due to all pairs of charges. If the charges are labeled 1, 2, 3, …, then the potential energy is

U_{elec} = \sum_{All pairs}^{} \frac{K q_{i} q_{j}}{r_{i j}}

(25.15)

where $r_{i j}$ is the distance between $q_{i}$ and $q_{j}$ . Be sure to include each pair only once.

NOTE For energy-conservation problems, it’s necessary to calculate only the potential energy for those pairs of charges for which the distance $r_{i j}$ changes. The potential energy of any pair that doesn’t move is an additive constant with no physical consequences.

EXAMPLE 25.4 Launching an electron

Three electrons are spaced 1.0 mm apart along a vertical line. The outer two electrons are fixed in position.

Is the center electron at a point of stable or unstable equilibrium?
If the center electron is displaced horizontally by a small distance, what will its speed be when it is very far away?

MODEL Energy is conserved. The outer two electrons don’t move, so we don’t need to include the potential energy of their interaction.

VISUALIZE FIGURE 25.12 shows the situation.

The figure illustrates the before and the after position of three electrons.

FIGURE 25.12 Full Alternative Text

SOLVE

The center electron is in equilibrium exactly in the center because the two electric forces on it balance. But if it moves a little to the right or left, no matter how little, then the horizontal components of the forces from both outer electrons will push the center electron farther away. This is an unstable equilibrium for horizontal displacements, like being on the top of a hill.
A small displacement will cause the electron to move away. If the displacement is only infinitesimal, the initial conditions are ${(r_{12})}_{i} = {(r_{23})}_{i} = 1.0 mm$ and $v_{i} = 0$ . “Far away” is interpreted as $r_{f} \to \infty$ , where $U_{f} \approx 0$ . There are now two terms in the potential energy, so conservation of energy $K_{f} + U_{f} = K_{i} + U_{i}$ gives

$\begin{array}{l} \frac{1}{2} m v_{f}^{2} + 0 + 0 & = 0 + [\frac{K q_{1} q_{2}}{{(r_{12})}_{i}} + \frac{K q_{2} q_{3}}{{(r_{23})}_{i}}] \\ = [\frac{K e^{2}}{{(r_{12})}_{i}} + \frac{K e^{2}}{{(r_{23})}_{i}}] \end{array}$

This is easily solved to give

$v_{f} = \sqrt{\frac{2}{m} [\frac{K e^{2}}{{(r_{12})}_{i}} + \frac{K e^{2}}{{(r_{23})}_{i}}]} = 1000 m / s$