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?

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 \( \mathbf{F}_{12} \), \( \mathbf{F}_{12} \) is given by Eq. (1.3) even though other charges are present. Thus, \[ \mathbf{F}_{12}=\frac{1}{4\pi\varepsilon_{0}} \frac{q_{1}q_{2}}{r_{12}^{2}}\hat{\mathbf{r}}_{12} \] In the same way, the force on \( q_{1} \) due to \( q_{3} \), denoted by \( \mathbf{F}_{13} \), is given by \[ \mathbf{F}_{13}=\frac{1}{4\pi\varepsilon_{0}} \frac{q_{1}q_{3}}{r_{13}^{2}}\hat{\mathbf{r}}_{13} \] 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 \( \mathbf{F}_{1} \) on \( q_{1} \) due to the two charges \( q_{2} \) and \( q_{3} \) is given as

\[ \mathbf{F}_{1}=\mathbf{F}_{12}+\mathbf{F}_{13}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r_{12}^{2}} \hat{\mathbf{r}}_{12}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{3}}{r_{13}^{2}} \hat{\mathbf{r}}_{13} \] (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 \(\mathbf{F}_{1}\) on the charge \(q_{1}\), due to all other charges, is then given by the vector sum of the forces \(\mathbf{F}_{12}\), \(\mathbf{F}_{13}\), ..., \(\mathbf{F}_{1n}\):
i.e.,

\[ \begin{array}{l} \mathbf{F}_{1}=\mathbf{F}_{12}+\mathbf{F}_{13}+\ldots+\mathbf{F}_{1 \mathrm{n}} \\ =\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q_{1} q_{2}}{r_{12}^{2}} \hat{\mathbf{r}}_{12}+\frac{q_{1} q_{3}}{r_{13}^{2}} \hat{\mathbf{r}}_{13}+\ldots+\frac{q_{1} q_{n}}{r_{1 n}^{2}} \hat{\mathbf{r}}_{1 n}\right. \\ =\frac{q_{1}}{4 \pi \varepsilon_{0}} \sum_{t=2}^{n} \frac{q_{t}}{r_{1 t}^{2}} \hat{\mathbf{r}}_{1 t} \end{array} \] (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.