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

Options

A. `(Q^(2))/((4piepsilon_(0)a^(2))`
B. `(-Q^(2))/((4piepsilon_(0)a^(2))`
C. Zero
D. `(Q^(2))/((2piepsilon_(0)a^(2))`

Answer

c

Explanation

The various forces acting at charge `+Q` are:
`vec(F)_(B)` force of attraction `= (1)/(4piepsilon_(0)).(Q^(2))/(a^(2))`
`vec(F)_(C )` is force of repulsion `= (1)/(4piepsilon_(0)).(Q^(2))/(a^(2))`
Resolving the force, we have `F_(B) sin 60^(@)` along `-y` direction and `F_(C ) sin 60^(@)` along `+y` direction. Two equal and opposite forces which cancel each other. Hence, net force is zero.