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Electric field of a line charge with the divergence theorem

  1. Nov 8, 2012 #1

    on page 63 of David J. Griffiths' "Introduction to Electrodynamics" he calculates the electric field at a point z above a line charge (with a finite length L) using the electric field in integral form.
    [itex]E_z = \frac{1}{4 \pi \epsilon_0} \int_{0}^{L} \frac{2 \lambda z}{\sqrt{(z^2 + x^2)^3}} dx = \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda L}{z \sqrt{z^2 + L^2}}[/itex]
    whereas [itex] \lambda = \frac{Q}{L}[/itex]
    Basically it's a two-dimensional system with a horizontal x-axis and a vertical z-axis, the charges go from -L to L on the x-axis and we look at the electric field a distance z above the line (i.e. on the z-axis).

    That's all fine and dandy but I have some serious troubles trying to reproduce that same result with the corresponding Maxwell equation in differential form and the divergence theorem.
    This is what I got so far:

    \vec{\nabla} \vec{E} = \frac{\rho}{\epsilon_0} \\
    \int \vec{\nabla} \vec{E} dV = \int \frac{\rho}{\epsilon_0} dV \\
    \int \vec{E} d\vec{A} = \int \frac{\lambda}{\epsilon_0} dl \\
    whereas I used [itex] \rho dV \propto \lambda dl [/itex]

    For the left side I use cylindrical coordinates and get:
    \vec{E} \hat{x} = \frac{1}{2 \pi \epsilon_0 x z} \int {\lambda} dl

    Since λ is constant I can pull it out of the integral and when I integrate I get as a final result:
    [itex]\vec{E} \hat{x} = \frac{2 \lambda L}{4 \pi \epsilon_0 x z}[/itex]

    Now this is a completely different result than what I get when I use the formula for the electric field in the integral form. One of the problems that this happens is that the integral form actually has the vector difference between the position of the charge and the point at which you want to calculate the electric field, i.e. [itex] \int \lambda \frac{\vec{r}-\vec{r'}}{\| \vec{r} - \vec{r'}\|^3} dl [/itex] whereas this is not the case in the Maxwell equation.

    What am I doing wrong? Why cannot I reproduce the same result?
  2. jcsd
  3. Nov 8, 2012 #2


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    How? Your system has no symmetry which could be used in the integral.
  4. Nov 8, 2012 #3
    I am not sure I understand what you mean.

    Though I think I have made a mistake.
    I wanted to integrate using the surface area for cylindrical coordinates, i.e.:
    [itex] \int \vec{E} \; d\vec{A} = \int \vec{E} \; \hat{r} \; dr \; d\phi \; dx = \vec{E} \; \vec{r} \; ln(r) \; 2 \pi \cdot 2 L [/itex]

    whereas r is the radius of the Gaussian cylinder I have put around the line charge.
    Although I just realised that this is actually the integral for the volume of a cylinder. I am not really sure how to integrate over the surface but I don't think that will solve the problem.

    I still don't understand why it's not possible to use the divergence theorem to solve this problem. :/
  5. Nov 8, 2012 #4
    It's not that you can't use the divergence theorem, as it holds no matter the case. You cannot use cylindrical symmetry to simplify the system because the E field in the system will not be cylindrically symmetric. To prove this to yourself, take the shape of the field when z>>L.

    Even as a volume integral, this is incorrect, as [itex]\hat{r}[/itex] is a unit vector, and I don't know where you are getting the ln(r) from. The area element [itex]d\vec{A} = \vec{r} \; d\phi dx [/itex] is proportional to and in the direction of [itex]\vec{r}[/itex]. To get the surface area, the area element is integrated at constant radius r=R. However, [itex]\vec{E} \cdot \hat{r}[/itex] would be a function of r and x, so it cannot be pulled out of the surface integral.
  6. Nov 9, 2012 #5
    Yes, thank you!
    The I actually got the ln(r) from the unit vector, since
    [itex] \hat{r} = \frac{\vec{r}}{r} [/itex] and when integrating 1/r you get ln(r).
    You are right though, my calculations are a mess, too many mistakes. Sorry about that. :/

    At least I now know that this configuration does not permit cylindrical symmetry because of both ends of the cylinder, which seem to be the trouble makers.
    Something like that would work for an infinitely long cylinder, though.
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