Three charges of equal magnitude q is placed at the vertices of an equilateral triangle of side l. The force on a charge Q placed at the centroid of the triangle is

Explanation

In the given equilateral triangle ABC of sides of length l, if we draw a perpedicular AD to the side BC.
AD=`AC cos 30^(@) =(sqrt3//2)` and the distance AO of the entroid O from A is (2/3) AD= `(1//sqrt3)`l. By symmartry AO=BO=CO.
`"Force" F_(1) "on Q due to the charge 1 q at A" = (3)/(4piepsilon_(0))(Qq)/(l^(2))"along AO" `
`"Force" F_(2) "on Q due to the charge 1 q at B" = (3)/(4piepsilon_(0))(Qq)/(l^(2))"along BO" `
`"Force" F_(3) "on Q due to the charge 1 q at C" = (3)/(4piepsilon_(0))(Qq)/(l^(2))"along cO" `
`"The resultant of forces" F_(2) and F_(3) is = (3)/(4piepsilon_(0))(Qq)/(l^(2))"along OA," ` by the parallelogram law. Therefore, the total force on `Q= (3)/(4piepsilon_(0))(Qq)/(l^(2))(hatr-hatr)=0`, where `hatr` is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through 60° about O