Question

Three point charges are located on the x-axis at the following positions: Q1 = +4.00 μC is at x = 1.00 m, Q2 = +6.00 μC is at x = 0.00, and Q3 = -8.00 μC is at x = -1.00 m. What is the magnitude of the electric force on Q2? (k = 1/4πε0 = 8.99 × 109 N ∙ m2/C2)

Answer 1

Electrostatic force is given by:

F = k*Q1*Q2/R^2

Force is attractive if both charges have different signs and Force is repulsive if both forces have same sign.

Net force on charge q2 at origin will be

F12 is repulsive force towards left and F23 is attractive force towards left, So

F_net = -F12 - F23

F_net = - k*q1*q2/r12^2 - k*q2*q3/r23^2

r12 = distance between q1 and q2 = 1.00 m

r23 = distance between q2 and q3 = 1.00 m

So,

F_net = (-8.99*10^9*6.00*10^-6*4.00*10^-6/1.00^2) - (8.99*10^9*6.00*10^-6*8.00*10^-6/1.00^2)

F_net = -0.647 N

Magnitude of net force = |F_net| = 0.647 N

Since F1 is negative, So direction of force is towards left.

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