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,

\[F_{BA}  =  F_{CA}  = F\left( \text{ say } \right)\]

 

 

 

The horizontal components of force cancel each other and the net force on the charge at A,

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

\[F' = 2F\sin\theta  \] 

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