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## Homework Statement

---- "+" ---- "+" ---- "-" ----

Given this configuration of charges where the plus and minus indicate the magnitude of charge and the distance between each charge on the line is 1 unit, solve for the point where E=0.

## Homework Equations

[itex]\Phi = \frac{q}{r}[/itex]

[itex]E = - \nabla \Phi [/itex]

[itex]F = [/itex][itex]q[/itex][itex]E[/itex]

## The Attempt at a Solution

I'm going to place my coordinate system on the middle charge, then use the principle of superposition to add together each individual potential, then take the negative gradient to solve for E. I'll label the charge to the left 1, the middle 2, and the right 3.

[itex]\Phi_1 = \frac{q}{x+1}[/itex]

[itex]\Phi_2 = \frac{q}{x}[/itex]

[itex]\Phi_3 = \frac{-q}{x-1}[/itex]

[itex]\Phi=\frac{q}{x+1}+\frac{q}{x}+\frac{-q}{x-1}[/itex]

[itex]E = -\frac{d}{dx}\Phi[/itex]

[itex]E = -\frac{d}{dx}(\frac{q}{x+1}+\frac{q}{x}+\frac{-q}{x-1})[/itex]

[itex]E = \frac{q}{(x+1)^2}+\frac{q}{x^2}+\frac{-q}{(x-1)^2}[/itex]

I made a plot of E(x) using Wolfram Alpha:

http://www.wolframalpha.com/input/?i=plot+1/(x+1)^2+++1/x^2+-+1/(x-1)^2,+x=-1+to+x=1

It looks like if I placed a test charge at x~0.5, it wouldn't experience a force, since E=0 there. x=0.5 is between the the positive charge in the middle and the negative charge on the right given I set x=0 on top of the middle charge.

Apparently the correct answer for E=0 is all the way past the negative charge to the right and I'm not sure what I did wrong. This is what the teacher said.

"If +1C is to the right from everything, when electron is pulling it

to the left and the protons are both pushing +1 to the right, the

protons have more charge (+2e), but the electron is closer, so it

could balance the protonic force, giving E=0."

Thanks for the help everyone.