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Electric potential linear charge

  1. Oct 21, 2014 #1
    1. The problem statement, all variables and given/known data
    A charge Q is uniformly distributed along a straight line of length a. Calculate V by integration, choosing the origin of the coordinates as the center of the charge.
    Next, expand this value of V up to terms in 1/r^2
    2. Relevant equations
    [itex] V = \frac{\lambda}{4 \pi \epsilon_0 r} [/itex]

    3. The attempt at a solution
    I let the line of charge sit along the z-axis, with its center at z=0, in the z-x plane
    I know the charge density [itex]\lambda=\frac{Q}{a} [/itex]

    [itex] V = \frac{Q}{4\pi a \epsilon_0} \int_{-a/2}^{a/2} \frac{1}{\sqrt{z^2 + x^2}} \, dz[/itex]

    Solving the integral, this gives me the solution [itex] V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{x^2 + \frac{a^2}{4}}+ \frac{a}{2}}{\sqrt{x^2 + \frac{a^2}{4}}- \frac{a}{2}}[/itex]

    However, this is not the right answer. My book says the answer is [itex] V = \frac{Q}{4\pi a \epsilon_0} \ln \frac{\sqrt{r^2 - az + \frac{a^2}{4}}+ \frac{a}{2} -z}{\sqrt{r^2 + az + \frac{a^2}{4}}- \frac{a}{2} -z} [/itex]

    So, now I realize I solved for the potential only straight above the center point of the line, on the x-axis... So, I'm guessing my book wants a solution for the potential anywhere, and that is why their answer looks more complicated, with more variables...

    So, I'm having trouble generalizing my V formula to work for any point around the line of charge. I know I need to change [itex] \frac{1}{\sqrt{z^2 + x^2}} [/itex], but I'm not sure how exactly to change it..

    Also, I'm not sure what r and z correspond to in the book's solution - I'm not even sure how they set up their axis (no picture)
  2. jcsd
  3. Oct 22, 2014 #2
    You are correct in your assumptions, but you do need to find V for every point on the 3D plane.

    Try using the generic form for electric potential where you would replace 1/√z2 + x2
    with 1/√(x - x')2 + (y - y')2 + (z - z')2

    and think about which values would be zeroed due to the location of the rod.
  4. Oct 22, 2014 #3
    Yes I'm glad to see that suggestion, as it was something i was considering this morning. Because the line of charge is on the z-axis, x'=y'=0. So I just need to replace [itex] \frac{1}{\sqrt{x^2 + z^2}} [/itex] with [itex] \frac{1}{\sqrt{x^2 +y^2 + (z-z')^2}} [/itex] and solve this integral with dz'

    Then I suppose the book just replaced [itex] \sqrt{x^2 + y^2 + z^2} [/itex] with [itex] \sqrt{r^2} [/itex] in the answer?
  5. Oct 22, 2014 #4
    Yes, that is correct about √r2
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