Question

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

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

Answer 1

The correct option is C $zero$

The force at A due to charge +Q at C is outward from C and the force at A due to charge -Q at B is towards B as shown in figure. So the resultant is parallel to BC. Thus there will no force in a direction normal to BC.