Answer to Question #161812 in Electricity and Magnetism for moath

Question

In the figure particle 1 of charge q₁ = -5.34q and particle 2 of charge q₂ = +1.84q are fixed to an x axis. As a multiple of distance L, at what coordinate on the axis is the net electric field of the particles zero?

Answer 1

The field from the first particle points toward the particle since it's negative. The field from the other particle point outside of the particle as it is positive.

Assume the particles are separated by distance L and the coordinate where the field is 0 is at distance x from the first particle.

Calculate fields:

$$E_1=\frac{kq_1}{x^2},\space E_2=\frac{kq_2}{(L+x)^2}.\\\space\\ E_1+E_2=0,\\\space\\ \frac{kq_1}{x^2}=-\frac{kq_2}{(L+x)^2},\\\space\\ 2.9x^2+5.8Lx+2.9L^2=x^2,\\ 1.9x^2+5.8Lx+2.9L^2=0,\\ x=\frac{-5.8L\pm\sqrt{(-5.8L)^2-4·1.9·2.9L^2}}{2·1.9}=\\\space\\ =L(\pm0.89-1.5),\\ x_1=-0.61L,\\ x_2=-2.39 L.$$