Potential Energy of a system of Point Charges

[Tybstar]

1. The problem statement, all variables and given/known data

Two identical charges q are placed on the x axis, one at the origin and the other at x = 5 cm. A third charge -q is placed on the x axis so the potential energy of the three-charge system is the same as the potential energy at infinite separation. Its x coordinate is: (x=13)

2. Relevant equations

I know the correct answer to this one because I'm looking over a marked up test in preparation for the final.

I've been using the equations for the electric potential energy of a system of point charges:

U=W=k\frac{q_{1}q_{2}}{r}

3. The attempt at a solution

...and expanding that to the work of the entire system:

W=0=W_{12}+W_{23}+W_{13}

so,


W=0=k\frac{q_{1}q_{2}}{5}+k\frac{q_{2}q_{3}}{d}+k\ frac{q_{1}q_{3}}{5+d}

Since q_{3} is negative, simplify:

W=0=k\frac{q^{2}}{5}-k\frac{q^{2}}{d}-k\frac{q^{2}}{5+d}


I might be missing something simple in algebra, but I can't figure out how to simplify this down enough to solve for d.

Any ideas? Am I even on the right track here?

Thanks!

[chaoseverlasting]

kq^2 can be eliminated (as the right hand side is equal to zero), and this leaves you with:

\frac{1}{5} -\frac{1}{d} -\frac{1}{5+d}=0. Take the negative terms to one side and cross multiply to get a quadratic equation in d. Solve quadratic for d. There are two possible solutions as this is a quadratic.

[Tybstar]

I'm sorry, I forgot to mention that I tried doing a quadratic, but was unable to get the correct answer. Also, I think that the correct term should be \frac{1}{(d-5)}, not \frac{1}{5+d}.

[chaoseverlasting]

Yes. You're right. I didnt look at the whole post. The solution, however, is x=\frac{15}{2}+-\frac{5\sqrt{5}}{2}