# Question #70bb8

Dec 25, 2016

It is obvious from the given figure that three charge points are situated at thrre vertices of an isosceles triangle in which two equal sides of length $l$ is inclined at an angle ${120}^{\circ}$ and $- 2 q$ charge resides at the point of intersection of these two sides. So each of the other two angles of the isosceles triangle will be ${30}^{\circ}$ .The c.g of two $+ q$ charge will be at the mid point of the line segment joining other two vertices of the triangle and the distance of resultatant $+ 2 q$ charge at this middle point from $- 2 q$ charge will be $d = l \times \sin {30}^{\circ} = \frac{l}{2}$
$\mu = 2 q \times d = 2 q \times \frac{l}{2} = q l$