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Guass's Law for a charge distribution

  1. May 25, 2013 #1
    First, this is not a homework question, just something I've been confused about for some time. I understand how to use Guass's law in many ways but one thing I have always stumbled with is whether the E-field of a charge distribution should involve little r or big R in such an example:

    In finding the E-field of a long cylindrical charge distribution with radius R and a uniform charge density [itex]\rho[/itex]. If I want to find the E-field inside (r<R) such a distribution shouldn't the final answer contain a big R?

    Here is my work. Since we have cylindrical symmetry we can consider a guassian cylinder of radius r inside about the cylindrical axis. The total charge Q is given by [itex]Q=\rho V[/itex] where V is the volume of the entire cylinder correct? (This is where I get confused) Or just the Guassian cylinder?

    If the entire cylinder then the volume is [itex]V=\pi R^{2}L[/itex], where L is the length of the cylindrical volume. Then working out the rest of the problem we have:

    [itex]\oint EdA=E(2\pi r)L = \frac{Q}{\epsilon_{0}} = \frac{\rho \pi R^{2}L}{\epsilon_{0}}[/itex]
    [itex]E = \frac{\rho R^2}{2 \epsilon_{0} r}[/itex]

    If the Guassian cylinder then the volume is [itex]V=\pi r^{2}L[/itex], then the E-field is just:

    [itex]E = \frac{\rho r}{2 \epsilon_{0}}[/itex]

    Do you see my confusion? I need to understand this and I think it is just geometrical issue. . .
  2. jcsd
  3. May 25, 2013 #2


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    You need to be aware of a very important aspect of Gauss's Law - that the field at any point r depends ONLY on the charges enclosed by the surface of size r (under highly symmetric situation). The charges outside the surface do not contribute to the E-field.

    So in your example, you could have R = infinity if you wish, and you still have the same expression for the E-field, because it only cares about the amount of charges enclosed by your gaussian surface.

  4. May 25, 2013 #3


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    It's the Gaussian cylinder. Otherwise, there is no point to making the Gaussian surface in the first place.
  5. May 25, 2013 #4
    If you're confident you've done it right, you have to believe the results which you get!!
    Last edited: May 25, 2013
  6. May 31, 2013 #5
    According to the Gauss's Law, Q is the charge inside the gaussian surface. So, the 2nd answer involving 'r' is right.
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