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

Let the force acting on +q charge at O due to +q at A be `F_(1)`, +q at B be `F_(2)` and -q at C be `F_(3)`
Here `AO=BO=CO=(d)/(sqrt(3))`

In magnitude `F_(1)=F_(2)=F_(3)=(1)/(4pi in_(0))(3q^(2))/(d^(2))`
Resultant of `F_(1) and F_(2)` is `F_(4)=(1)/(4pi in_(0)) (3q^(2))/(d^(2))`
(as angle between `F_(1) and F_(2)` is `120^(0)`)
Direction of `F_(4)` is along the direction of `F_(3)`. Hence the resultant force on +q at O is
`F=F_(3)+F_(4)=(3q^(2))/(2pi in_(0)d^(2))`