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Homework Help: Finding potential at a point due to line of charge.

  1. Jul 1, 2008 #1


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

    Find the potential at the point A(0,0,2) due to a continuous line of charge with charge density 12nC/m on the line y=2.5, z=0. Zero potential is defined as that infinitely far away.

    In this question, I'll denote the charge density by p, the point for which the potential is measured as z, and the line of charge as y=c.

    2. Relevant equations

    [tex] V_A = - \int^A_\infty \vec{E} \cdot d\vec{L}[/tex]

    3. The attempt at a solution

    Ok, so the book says that the potential can be found by carrying a unit charge from infinity to the given point along any path we choose. So I'll just apply the formula:

    [tex] V_A = - \int^A_\infty \vec{E} \cdot d\vec{L}[/tex]

    Here's what I did:

    The differential electric field due to a charged segment dx is given by

    [tex]d\vec{E} = \frac{qdx(\vec{r} - \vec{r'})}{4\pi \varepsilon_0 |\vec{r} - \vec{r'}|^3}[/tex]
    [tex]\vec{r} - \vec{r'} = \left(\begin{array}{cc}-x\\-c\\z\end{array} \right) [/tex]
    [tex]|\vec{r} - \vec{r'}|^3 = (x^2+c^2+z^2)^{3/2}[/tex]
    [tex]\frac{p}{4\pi \varepsilon_0} \frac{dx}{(x^2+c^2+z^2)^{3/2}} \left(\begin{array}{cc}-x\\-c\\z\end{array} \right)[/tex]

    By symmetrical considerations, E in x direction is 0, so I'll let x above = 0.
    Integrating [tex]d\vec{E} [/tex] over dx from [tex]-\infty \ \mbox{to} \ \infty[/tex] gives E:

    [tex]\vec{E} = \frac{p}{4\pi \varepsilon_0} \int^\infty_{-\infty} -\frac{c \ dx}{(x^2+c^2+z^2)^{3/2}} \mbox{j} + \frac{z \ dx}{(x^2 + c^2 + z^2)^{3/2}} \mbox{k} [/tex]

    This works out, using the online integrator, as

    [tex]\vec{E} = \frac{p}{4 \pi \varepsilon_0} \cdot \left[ -\frac{cx}{(z^2+c^2)\sqrt{z^2+c^2+x^2}} \right]^\infty_{-\infty} \mbox{j} + \left[ -\frac{zx}{(z^2+c^2)\sqrt{z^2+c^2+x^2}} \right]^\infty_{-\infty} \mbox{k}[/tex]

    Taking the limits yield:

    [tex]\vec{E} = \frac{p}{4\pi \varepsilon_0} \cdot \frac{2}{c^2 + z^2} \left(\begin{array}{cc}0\\-c\\z\end{array} \right)[/tex]

    Now applying the formula yields:

    [tex]V_A = -\int^A_\infty \vec{E} \cdot d\vec{L}[/tex]
    [tex]= - \int^A_\infty \frac{p}{4 \pi \varepsilon_0} \frac{2}{z^2+c^2} \left(\begin{array}-0\\-c\\z\end{array} \right) \cdot \left(\begin{array}{cc}dx\\dy\\dz\end{array}\right)[/tex]
    [tex]=-\frac{p}{4 \pi \varepsilon_0} \left[ \int^0_\infty -\frac{2c}{z^2+c^2} dy + \int^z_\infty \frac{2z}{z^2 + c^2} dz \right][/tex]

    This gives:
    [tex] -\frac{p}{4 \pi \varepsilon_0} ln|z^2 + c^2|[/tex]

    Substituting the values of c,z and p gives the answer as -251.3V whereas the actual answer is 67.4V. Where's my mistake?
  2. jcsd
  3. Jul 1, 2008 #2
    I'm confused. Is this a infinite straight line of constant charge density? If this is the case, then the electric field varies inversely with distance from the line. So the potential should be infinite...
  4. Jul 1, 2008 #3


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    The question states:

    So I assumed that the line would be infinite. But somehow the answer is finite.
  5. Sep 28, 2008 #4


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    Hi guys, really sorry to have to drag up this thread. But I came by this question while doing my revision and it turns out that so long as 0 potential is defined as at infinity, the answer would be indeterminate. I would post the working but it's a little tedious and instead of the method above, I simply evaluated the potential directly. I also found this:
    So I'd like the experts to confirm if this is indeed correct.
    Last edited by a moderator: Apr 23, 2017
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