Question

Three equal charges, 2.0 × 10⁻⁶ C each, are held at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the other two.

Answer 1

Since all the charges are of equal magnitude, the force on the charge at A due to the charges at B and C will be of equal magnitude. (As shown in the figure)

That is, $F_{\mathrm{BA}}=F_{\mathrm{CA}}=F\left(\mathrm{say}\right)$


The horizontal components of force cancel each other and the net force on the charge at A,
$$\begin{gathered} F\text{'}=F_{\mathrm{BA}}\sin \theta +F_{\mathrm{CA}}\sin \theta \\ \mathrm{F}\text{'}=2F\sin \theta \end{gathered}$$
Given: r = 5 cm =0.05 m
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$$\begin{gathered} F\text{'}=\frac{2\times 9\times {10}^9\times {\left(2\times {10}^{-6}\right)}^2\times {\displaystyle \sin 60°}}{{\left(0.05\right)}^2} \\ F\text{'}=\frac{2\times 9\times {10}^9\times {\left(2\times {10}^{-6}\right)}^2\times {\displaystyle \frac{\sqrt{3}}{2}}}{{\left(0.05\right)}^2} \end{gathered}$$
F' = 24.9 N