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 at the vertex A in a direction perpendicular 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

The correct option is C zero


The various forces acting at charge +Q are $F_B$ force of attraction
$=\frac{1}{4\pi {\epsilon }_0}.\frac{Q^2}{a^2}$

At $C$ the force there will be repulsive force.
So, $F_c$ is force of repulsion $=-\frac{1}{4\pi {\epsilon }_0}.\frac{Q^2}{a^2}$

Resolving the force, we have $F_B\sin {60}^{\circ }$ and $F_C\sin {60}^{\circ }$ are equal and opposite forces, will cancel each other. Hence net force in a direction perpedicular to BC is zero.