Example

Question:

Consider the charges \( q \), \( q \), and \( -q \) placed at the vertices of an equilateral triangle, as shown in Fig. 1.7. What is the force on each charge?

Solution:

The forces acting on charge \( q \) at A due to charges \( q \) at B and \( -q \) at C are \( \mathbf{F}_{12} \) along BA and \( \mathbf{F}_{13} \) along AC respectively. By the parallelogram law, the total force \( \mathbf{F}_1 \) on the charge \( q \) at A is given by
\[ \mathbf{F}_1 = F\, \mathbf{\hat{r}}_1 \] where \( \mathbf{\hat{r}}_1 \) is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude \[ F = \frac{q^2}{4\pi\varepsilon_0 l^2} \] The total force \( \mathbf{F}_2 \) on charge \( q \) at B is thus \[ \mathbf{F}_2 = F\, \mathbf{\hat{r}}_2 \] where \( \mathbf{\hat{r}}_2 \) is a unit vector along AC.
Similarly, the total force on charge \( -q \) at C is \[ \mathbf{F}_3 = \sqrt{3}\,F\, \mathbf{\hat{n}} \] where \( \mathbf{\hat{n}} \) is the unit vector along the direction bisecting the \( \angle \text{BCA} \).
The sum of the forces on the three charges is zero, i.e., \[ \mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = 0 \] This follows from Coulomb’s law and Newton’s third law.