Question

A positive point charge Q₁ = 1.8 ? 10^-5 C is fixed at the origin of coordinates, and a negative charge Q₂ = -3.6 ? 10^-6 C is fixed to the x axis at x = +2.0 m. Find the location of the place(s) along the x axis where the electric field due to these two charges is zero. (Enter the locations in order from -x to +x. If there is only 1 location, enter 0 as the second answer.)
x = ______ m
x = ______ m

Answer 1

he positive charge at the origin is so much greater than the negative, then the field lines going from it are going to be pretty great, despite the farther distance from it. So you can conclude at some point out there beyond x =2, that the Electric field from the positive charge is going to cancel the electric field going to the negative. And I'll tell you that it should be pretty far out there probably because that +charge is a 1000x more greater than the negative charge. But remember too that distance is inversely proportional, so I guess it won't be "that" far out. But far nonetheless.

So call the electric field due to the + charge E1 and the field due to - charge E2. x, where the Enet = 0, will be the distance from the + charge at the origin.

Enet = 0 = E1 - E2
E1 = E2
kQ1 / x^2 = kQ2 / (x +2)^2

And then of course k = 9x10^9, Q1 = 1.8x10^-5 and Q2 = -3.5x10^-6.After k gets cancelled

Q1 / x^2=Q2 / (x +2)^2

.8x10^-5/x*2=-3.5x10^-6/(2+2)^2

220=16 x^2

x=3.75 m