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 $B C$ is

• A. $\frac{Q^{2}}{\left(4 \pi \varepsilon_{0} a^{2}\right)}$
• B. $\frac{-Q^{2}}{\left(4 \pi \varepsilon_{0} a^{2}\right)}$
• C. zero
• D. $\frac{Q^{2}}{\left(2 \pi \varepsilon_{0} a^{2}\right)}$

Answer 1

Answer: (C)

The various forces acting at charge $+Q$ are:
$\vec{F}_{B}$ (force of attraction) $=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{a^{2}}$
$\vec{F}_{C}$ (force of repulsion) $=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{2}}{a^{2}}$
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.