Three charges are placed at the vertices of an equilateral triangle of side l. Each of the charges is q. Find out the force on a charge Q placed at the center of mass of the triangle.

Explanation

The centre of mass O is equidistant from the vertices A,B ,C of the equilateral triangle. (Fig.)

So, each of the charge q will exert equal force F on the charge Q at O. i.e, `F_(A) F_(B) = F_(C)=F`
Now, two equal forces `F_(B)` and `F_(C)` act at O along OE and OF, respectively. So their resultant F' will act along OA, which is inclined at an angle of `60^(@)` with either OE or OF.
`therefore F' = F_(B) cos 60^(@) + F_(C) cos60^(@) =F.1/2 + F.1/2 =F`
As the force `F_(A)` of magnitude F, acts along OD, i.e., opposite to OA, the resultant of F' and `F_(A)` will be zero.
`therefore vecF_(A) + vecF_(B) + vecF_(C) = vecF_(A) + vecF' =-vecF + vecF=0`
So, any charge Q placed at the centre of mass of a triangle will experiences no force.