Question

A test charge of +2 μ C is placed halfway between a charge of +6 μ C and another of +4 μ C separated by 10 cm. (a) What is the magnitude of the force on the test charge? (b) What is the direction of this force (away from or toward the +6 μ C charge)?

Answer 1

Consider q1 = 6 uC = 6 x 10^-6 C

q2 = 4 uC = 4 x 10^-6 C

and q = 2 uC = 2 x 10^-6 C

r = 10/2 = 5 cm = 0.05 m

(a) Force on q due to charge q1 -

F1 = k*q1*q / r^2 = (9 x 10^9 x 6 x 10^-6 x 2 x 10^-6) / 0.05^2 = 43.2 N (repulsive in nature)

Now, force on q due to charge q2 -

F2 = k*q2*q / r^2 = (9 x 10^9 x 4 x 10^-6 x 2 x 10^-6) / 0.05^2 = 28.8 N (repulsive in nature)

Therefore the net force on the test charge = F1 - F2 = 43.2 - 28.8 = 14.4 N

(b) Both the forces are repulsive in nature because all the charges are positive.

Charge 6 uC is larger than the charge +4 uC.

So, the direction of the net force (14.4 N) shall be away from +6 uC charge.