A 5.0 nC point charge sits at x = 0. At the same time, a 3500N/C uniform electric field (created by distant source charges) points in the positive x-direction.
At what point along the x-axis, if any, would a proton experience no net force?
At what point along the x-axis, if any, would an electron experience no net force?
Ok, so. What you want to do here is add the two acting forces on the second point charge (being the proton or the electron) and make that sum equal to zero, meaning no net force.
The force from the 5.0 nC point charge is the equation F = k* (q1)(q2)/r^2
The force from the distant source charge is E = F/q, which becomes F = E*q. q in this case is the charge of your second point (1.6*10^-19C for proton, -1.6*10^-19C for electron)
Adding these two equations together gives you E*q + k * (q1 * q2)/r^2 = 0
Plug in what you have, for the proton, it is: 3500N/C*(1.6*10^-19C) + ((9*10^9Nm^2/C^2) * (5*10^-9C) * (1.6*10^-19C))/r^2
Solve for r. Turns out, the answer is the same for the electron as it is for the proton. Look at this logically:
If you have a point charge with a field going in the positive x-direction, then it is a positively charged point. The 5.0nC point charge is also positive. Your answer will be on the negative x-axis, because with a proton, it is being equally repelled by both positive points. With the electron, it is being equally ATTRACTED to both points.
Oh, and by cancelling out your units, it shows that r is in meters (m).Source(s): I’m in physics 122 right now. Took a while to figure this out. Thought I’d help!