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## Homework Statement

A point charge q

_{1}= 4.10 nC is placed at the origin, and a second point charge q

_{2}= -2.95 nC is placed on the x-axis at x =+ 20.0 cm. A third point charge q

_{3}= 2.00 nC is to be placed on the x-axis between q

_{1}and q

_{2}. (Take as zero the potential energy of the three charges when they are infinitely far apart.)

What is the potential energy of the system of the three charges if q

_{3}is placed at x = + 11.0 cm?

Where should q

_{3}be placed between q

_{1}and q

_{2}to make the potential energy of the system equal to zero?

## Homework Equations

[tex]V_a-V_b = \int^b_aFdr[/tex]

[tex]U=qV[/tex]

[tex]k=\frac{1}{4\pi\epsilon_0}[/tex]

## The Attempt at a Solution

I'm thinking the problem I'm having at the moment is conceptual. I can't find an example of such a problem in the textbook, and haven't quite wrapped my mind completely around the subject...

Anyway, here's what I tried.

[tex]V-0=\int^\infty_xFdr=\int^\infty_xk\frac{q}{r^2}dr=kq(-\frac{1}{r})^\infty_x=kq(-\frac{1}{x})=-k\frac{q}{x}[/tex]

I used the result from that as follows:

[tex]V_1=-k\frac{q}{x}=-k\frac{4*10^{-9}\ C}{1.1*10^{-1}\ m}=-326.9747\ V[/tex]

[tex]V_2=-k\frac{q}{x}=-k\frac{-2.95*10^{-9}\ C}{9*10^{-2}\ m}=294.731\ V[/tex]

[tex](V_1+V_2)q_3=-6.44874*10^{-8}\ J[/tex]

Naturally this is wrong. To make it easier to read, I attempted to find the potential between q

_{1}and q

_{3}, as well as q

_{2}and q

_{3}.

Do I just need to add in a calculation for the potential between q

_{1}and q

_{2}? Or, is there something else I need to do?