1.7 FORCES BETWEEN MULTIPLE CHARGES

The mutual electric force between two charges is given by Coulomb’s law. How to calculate the force on a charge where there are not one but several charges around? Consider a system of n stationary charges $q_{1}, q_{2}, q_{3}, . . ., q_{n}$ in vacuum. What is the force on $q_{1}$ due to $q_{2}, q_{3}, . . ., q_{n}$ ? Coulomb’s law is not enough to answer this question. Recall that forces of mechanical origin add according to the parallelogram law of addition. Is the same true for forces of electrostatic origin?

1.8

A system of
(a) three charges
(b) multiple charges.

Experimentally, it is verified that force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges. This is termed as the principle of superposition.

To better understand the concept, consider a system of three charges $q_{1}, q_{2}$ and $q_{3}$ , as shown in Fig. 1.8(a). The force on one charge, say $q_{1}$ , due to two other charges $q_{2}, q_{3}$ can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on $q_{1}$ due to $q_{2}$ is denoted by $F_{12}$ , $F_{12}$ is given by Eq. (1.3) even though other charges are present. Thus, $F_{12} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{r}}_{12}$ In the same way, the force on $q_{1}$ due to $q_{3}$ , denoted by $F_{13}$ , is given by $F_{13} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{3}}{r_{13}^{2}} {\hat{r}}_{13}$ which again is the Coulomb force on $q_{1}$ due to $q_{3}$ , even though other charge $q_{2}$ is present.

Thus the total force $F_{1}$ on $q_{1}$ due to the two charges $q_{2}$ and $q_{3}$ is given as

$F_{1} = F_{12} + F_{13} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{r}}_{12} + \frac{1}{4 π ε_{0}} \frac{q_{1} q_{3}}{r_{13}^{2}} {\hat{r}}_{13}$ (1.4)

The above calculation of force can be generalised to a system of charges more than three, as shown in Fig. 1.8(b).

The principle of superposition says that in a system of charges $q_{1}$ , $q_{2}$ , ..., $q_{n}$ , the force on $q_{1}$ due to $q_{2}$ is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges $q_{3}$ , $q_{4}$ , ..., $q_{n}$ . The total force $F_{1}$ on the charge $q_{1}$ , due to all other charges, is then given by the vector sum of the forces $F_{12}$ , $F_{13}$ , ..., $F_{1 n}$ :
i.e.,

$\begin{array}{l} F_{1} = F_{12} + F_{13} + \dots + F_{1 n} \\ = \frac{1}{4 π ε_{0}} [\frac{q_{1} q_{2}}{r_{12}^{2}} {\hat{r}}_{12} + \frac{q_{1} q_{3}}{r_{13}^{2}} {\hat{r}}_{13} + \dots + \frac{q_{1} q_{n}}{r_{1 n}^{2}} {\hat{r}}_{1 n} \\ = \frac{q_{1}}{4 π ε_{0}} \sum_{t = 2}^{n} \frac{q_{t}}{r_{1 t}^{2}} {\hat{r}}_{1 t} \end{array}$ (1.5)

The vector sum is obtained as usual by the parallelogram law of addition of vectors. All of electrostatics is basically a consequence of Coulomb’s law and the superposition principle.

Example 1.6

Consider three charges $q_{1}$ , $q_{2}$ , $q_{3}$ each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. 1.9?

1.9

VIEW SOLUTION

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side BC, AD = AC cos $30^{0}$ = ( $\sqrt{3}$ /2)l and the distance AO of the centroid O from A is (2/3) AD = (1/ $\sqrt{3}$ ) l. By symmatry AO = BO = CO.

Thus,
Force $F_{1}$ on Q due to charge q at A = $\frac{3}{4 π ε_{0}} \frac{Q q}{l^{2}}$ along AO

Force $F_{2}$ on Q due to charge q at B = $\frac{3}{4 π ε_{0}} \frac{Q q}{l^{2}}$ along BO

Force $F_{3}$ on Q due to charge q at C = $\frac{3}{4 π ε_{0}} \frac{Q q}{l^{2}}$ along CO

The Resultant of forces $F_{2}$ , $F_{3}$ is $\frac{3}{4 π ε_{0}} \frac{Q q}{l^{2}}$ along OA, byt the parallelogram law. Therefore, the total force on Q= $\frac{3}{4 π ε_{0}} \frac{Q q}{l^{2}} (\hat{r} - \hat{r}) = 0$ , where $\hat{r}$ is the unit vector along OA.

It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60° about O.

Example 1.7

Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Fig. 1.10. What is the force on each charge?

1.10

VIEW SOLUTION

Solution The forces acting on charge q at A due to charges q at B and –q at C are $F_{12}$ along BA and $F_{13}$ along AC respectively, as shown in Fig. 1.10. By the parallelogram law, the total force $F_{1}$ on the charge q at A is given by
$F_{1}$ = F ${\hat{r}}_{1}$ where ${\hat{r}}_{1}$ is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude $F = \frac{q^{2}}{4 π ε_{0} l^{2}}$
The total force $F_{2}$ on charge q at B is thus $F_{2}$ = F ${\hat{r}}_{2}$ , where ${\hat{r}}_{2}$ is a unit vector along AC.

Similarly the total force on charge –q at C is $F_{3} = \sqrt{3} F \hat{n}$ , where $\hat{n}$ is the unit vector along the direction bisecting the $∠$ BCA. It is interesting to see that the sum of the forces on the three charges is zero, i.e., $F_{1} + F_{2} + F_{3} = 0$ The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law. The proof is left to you as an exercise.