Question

Three point charges are arranged on a line. Charge q3=+5.00nC and is at the origin. Charge q2=−3.00nC and is at x= 4.00 cm . Charge q1 is at x= 1.00 cm . What is q1 (magnitude and sign) if the net force on q3 is zero?

Answer 1

Given that net force on q3 is zero,

F3 = F13 - F23 = 0

F13 = F23

Now electrostatic force is given by:

F = kQ1Q2/R^2

F13 = kq1q3/r13^2

F23 = kq2q3/r23^2

So,

kq1q3/r13^2 = kq2q3/r23^2

q1/r13^2 = q2/r23^2

q1 = q2*(r13/r23)^2

q2 = 3 nC

r13 = 1 cm

r23 = 4 cm

So,

q1 = 3*(1/4)^2

q1 = 0.188 nC

since q2 and q3 are of opposite sign so force between them is attractive, which means force between q1 and q3 will be repulsive so q1 should be positive charge

q1 = +0.188 nC