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Potential energy of a system of three charges.

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations
    [tex]V_a-V_b = \int^b_aFdr[/tex]
    [tex]U=qV[/tex]
    [tex]k=\frac{1}{4\pi\epsilon_0}[/tex]

    3. 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 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?
     
  2. jcsd
  3. Sep 16, 2011 #2

    vela

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    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 have[tex]V_a - V_b = -\int_b^a \mathbf{E}\cdot d\mathbf{r}[/tex]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 [tex]V = \int_x^\infty \frac{kq}{r^2}\,dr[/tex]If 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 [tex]U=\frac{kq_1q_2}{r}[/tex]which is probably one of the formulas in your textbook.
     
    Last edited: Sep 16, 2011
  4. Sep 16, 2011 #3
    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...
    [tex]U=\frac{kq_1q_2}{r}[/tex] which makes sense, since [tex]V=\frac{kq_1}{r}[/tex] seems to be what I should have calculated in the problem.
    Is this what you meant?

    My new calculations are using
    [tex]U=\frac{kq_iq_j}{r_{ij}}[/tex]
    Which I utilized as follows:
    [tex]U=k\sum_{i<j}\frac{q_iq_j}{r_{ij}}[/tex]
    Which becomes:
    [tex]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[/tex]
    Which is apparently wrong...
     
    Last edited: Sep 16, 2011
  5. Sep 16, 2011 #4
    Nevermind. I had the figures right, but somehow calculated it wrong.
    The problem has been solved now. :)
     
  6. Sep 16, 2011 #5

    vela

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