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.

#### Solution

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,

\[F' = F_{BA} \sin\theta + F_{CA} \sin\theta\]

\[F' = 2F\sin\theta \]

Given: r = 5 cm =0.05 m

By Coulomb's Law, force,

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]

\[F' = \frac{2 \times 9 \times {10}^9 \times \left( 2 \times {10}^{- 6} \right)^2 \times \sin60^\circ}{\left( 0 . 05 \right)^2}\]

\[F' = \frac{2 \times 9 \times {10}^9 \times \left( 2 \times {10}^{- 6} \right)^2 \times \frac{\sqrt{3}}{2}}{\left( 0 . 05 \right)^2}\]

F' = 24.9 N