Question

Three charges lie along the x - axis. A positive charge q1 = 10 𝝁C is at x = 1.00 m, and a negative charge q2 = – 2.0 𝝁C is at the origin. Where must a positive charge q3 be placed on the x - axis so that the resultant force on it is zero?

Answer 1

$$F_{13}=F_{23}$$

$$k\frac{|q_1||q_3|}{(1.0+d)^2}=k\frac{|q_2||q_3|}{d^2}$$

$$\frac{10*10^{-6}}{(1.0+d)^2}=\frac{2.0*10^{-6}}{d^2}$$

$$\sqrt{5}d=1+d$$

$$d=\frac{1}{\sqrt{5}-1}=0.81\:\rm m$$

The third charge should be placed at $x=-0.81\:\rm m$.