Consider three charges `q_(1), q_(2), q_(3)` each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. ?

Explanation

In the given equilateral triangle ABC of sides of length l if we draw a perpendicualre AD to the side BC
AD=AC cos `30^(@) =sqrt(3)//2` l and the distance AO of the centroid o from A is `(2/3)` AD=`(1/sqrt(3))` l by symmatry AO=BO=CO
force `F_(1)` and Q due to charge q at A=`(3)/(4piepsilon_(0))(Qq)/(l^(2))` along AO
force `F_(2)` and Q due to charge q at B=`(3)/(4piepsilon_(0))(Qq)/(l^(2))` along AO
force `F_(3)` and Q due to charge q at C=`(3)/(4piepsilon_(0))(Qq)/(l^(2))` along AO
parallelogram law therefore the toal force on Q =`(3)/(4piepsilon_(0))(Qq)/(4piepsilon_(0))(Qq)/(l^(2))(vecr-vecr)`
=0 where `vecr` is the unit vector along OA
it is clear alos by symmetry that the three forces will sum to zero consider what would happen if the system was rotated through `60^(@)` about O