Question

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charge is 6.0 mu C, and the lengths of the sides of the triangle are 3.6 cm. Calculate the magnitude of the net force that each charge experiences. F_A = F_B = F_C =

Answer 1

The magnitude of the electrostatic force between two charge :
F = kQ1*q2/r^2 ---> Q1 = Q2 =Q
= k Q^2/r^2
= 9 x 10^9 * (6 x 10^-6)^2/9 x 10^-4 =360 N

Between the different forces acting on each other is an angle of 60°. Let the negative charge be at C.
Then the net force acting on C by the positive charges at A and B is
FC = √(360^2 + 360^2 +2*360*360*cos(60°)) (= length of long diagonal in a parallelogram)
FC = 623.53 N
and
FA = FB = √(360^2 + 360^2 - 2*360^2*cos(60°) (= length of short diagonal in a parallelogram)
= 360 N