Potential energy of a system of three charges.

[MSZShadow]

Homework Statement

A point charge q₁ = 4.10 nC is placed at the origin, and a second point charge q₂ = -2.95 nC is placed on the x-axis at x =+ 20.0 cm. A third point charge q₃ = 2.00 nC is to be placed on the x-axis between q₁ and q₂. (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 q₃ is placed at x = + 11.0 cm?

Where should q₃ be placed between q₁ and q₂ to make the potential energy of the system equal to zero?

Homework Equations

V_a-V_b = \int^b_aFdr
U=qV
k=\frac{1}{4\pi\epsilon_0}

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.
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}

I used the result from that as follows:
V_1=-k\frac{q}{x}=-k\frac{4*10^{-9}\ C}{1.1*10^{-1}\ m}=-326.9747\ V
V_2=-k\frac{q}{x}=-k\frac{-2.95*10^{-9}\ C}{9*10^{-2}\ m}=294.731\ V
(V_1+V_2)q_3=-6.44874*10^{-8}\ J

Naturally this is wrong. To make it easier to read, I attempted to find the potential between q₁ and q₃, as well as q₂ and q₃.
Do I just need to add in a calculation for the potential between q₁ and q₂? Or, is there something else I need to do?

 

[vela]

Yes, you need to take into account the potential energy between q₁ and q₂ as well.

Your equations are a bit odd. First, you should haveV_a - V_b = -\int_b^a \mathbf{E}\cdot d\mathbf{r}You used the letter F, which suggests you're integrating a force, but the potential difference is the integral of the electric field E.

Second, when you integrated, you got the sign wrong. You had V = \int_x^\infty \frac{kq}{r^2}\,drIf q>0, the integrand is positive, and you're integrating from some finite distance x out to infinity so dr>0 as well. The answer should be positive, but you ended up with a negative result.

If you fix that error, you find that the potential energy between two charges q₁ and q₂ separated by a distance r is U=\frac{kq_1q_2}{r}which is probably one of the formulas in your textbook.

 

[MSZShadow]

Ah. I had written that equation wrong in the problem statement/etc. Lovely. It looks like I had calculated it by integrating E, as opposed to F anyway.

And, I see where my sign went wrong. Products of a tired mind...

Thank you for clearing that up. :)

Edit: My textbook says that the potential energy between two charges q₁ and q₂ separated by a distance r is...
U=\frac{kq_1q_2}{r} which makes sense, since V=\frac{kq_1}{r} seems to be what I should have calculated in the problem.
Is this what you meant?

My new calculations are using
U=\frac{kq_iq_j}{r_{ij}}
Which I utilized as follows:
U=k\sum_{i<j}\frac{q_iq_j}{r_{ij}}
Which becomes:
U=\frac{1}{4\pi\epsilon_0}(\frac{4.1*10^{-9}\ C*-2.95*10^{-9}\ C}{0.2\ m}+\frac{4.1*10^{-9}\ C*2*10^{-9}\ C}{0.11\ m}+\frac{-2.95*10^{-9}\ C*2*10^{-9}}{0.09\ m})=498.636441\ J
Which is apparently wrong...

 

[MSZShadow]

Nevermind. I had the figures right, but somehow calculated it wrong.
The problem has been solved now. :)

 

[vela]

Edit: My textbook says that the potential energy between two charges q₁ and q₂ separated by a distance r is...
U=\frac{kq_1q_2}{r} which makes sense, since V=\frac{kq_1}{r} seems to be what I should have calculated in the problem.
Is this what you meant?

Yes, sorry. It should be r, not r². I fixed my earlier post.