Answer to Question #225185 in Electricity and Magnetism for Olavi

Question

A charge q1= 300 micro coulomb is at(0,0) and q2= -5/3q1 is at 20 cm mark on the positive x axis . We also have a charge q3= 8/5q2 located at some unknown point on the x axis . Where is we if the net force on q1 is 7.00 N pushing q1 to the left

Answer 1

First, we have to consider that the two forces exerted on charge 1 since charge 2 and charge 3 are both negative, we have two attractive forces and q₃ has to be located then at x<0 because F₁₂ is directed to the left as F_T. Then if we analyze the system we have:

$q_1=3\times10^{-4}\,C \\ q_2=-5\times10^{-4}\,C \\ q_3=-8\times10^{-4}\,C$

$F_T=-F_{12}+F_{13}=-7\,N$

$F_T=-\dfrac{kq_1q_2}{r^2_{12}}+\dfrac{kq_1q_3}{r^2_{13}} \\ F_T+\dfrac{kq_1q_2}{r^2_{12}}=\dfrac{kq_1q_3}{r^2_{13}} \\r_{13} =\sqrt{\dfrac{kq_1q_3}{F_T+\dfrac{kq_1q_2}{r^2_{12}}}} \\r_{13} =\sqrt{\dfrac{(9\times10^9Nm^2/C^2)(3\times10^{-4}\,C)(-8\times10^{-4}\,C)}{-7\,N+\dfrac{(9\times10^9Nm^2/C^2)(3\times10^{-4}\,C)(-5\times10^{-4}\,C)}{(0.2\,m)^2}}} \\ r_{13}=0.2529\,m$

The particle is located then at x= - 0.2529 m = - 25.29 cm