Question

Three charges are placed at the vertices of an equilateral triangle of side $a$ as shown in the following figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is

• A. $\frac{Q^2}{\left(4\pi {\epsilon }_0{a}^2\right)}$
• B. $\frac{-Q^2}{\left(4\pi {\epsilon }_0{a}^2\right)}$
• C. zero
• D. $\frac{Q^2}{\left(2\pi {\epsilon }_0{a}^2\right)}$

Answer 1

Answer: (C)

The various forces acting at charge $+Q$ are:
${\vec{F}}_B$ (force of attraction) $=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}$
${\vec{F}}_C$ (force of repulsion) $=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q^2}{a^2}$
Resolving the force, we have $F_B\sin 60^{\circ }$ along $-y$ direction and $F_C\sin 60^{\circ }$ along $+y$ direction. Two equal and opposite forces which cancel each other. Hence, net force is zero.