Charges of +2.0𝞵C, +3.0𝞵C and -8.0𝞵C are placed at the vertices of an equilateral triangle of side 10cm. calculate the magnitude of the force acting on the -8.0𝞵C charge due to the other two charges.

Answer 1

F vector = F1 vector + F2 vector
magnitude of F1 = k (8*10^-6)(3*101^-6)/ (0.1)^2 attracting
Magnitude of F2 = k (8*10^-6)(2*106-6) / (0.1)^2 attracting
third charge at origin
assume x in direction from third charge to second
then
Fx = F1 - F2 cos 60
Fy = F2 sin 60
magnitude = sqrt( Fx^2 + Fy^2)