Three equal charges each having charge +q are placed at the corners of an equilateral triangle of side  a. Then the Coulomb force experienced by one charge due to  rest of the two is

The force due to the two charges on the third charge be  $F_1$ and  $F_2$, then 

$F_1=F_2=\frac{1}{4\text{π}{\in }_0}\frac{q^2}{a^2}$

They act along the sides of triangle and angle between the forces is  ${60}^{\circ }$.
So the resultant force on the third charge is 

$\begin{array}{l}=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos {60}^{\circ }}\\ =\sqrt{F^2+F^2+2F^2\left(\frac{1}{2}\right)}\\ =\sqrt{3F^2}=\sqrt{3}F\\ F_R=\frac{1}{4\pi {\in }_0}\frac{\sqrt{3}{q}^2}{a^2}\end{array}$

Therefore, the correct answer is  (C)