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Simple electrostatics problem

  1. Feb 21, 2008 #1
    [SOLVED] simple electrostatics problem

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

    a cylinder, the inhomogenous E field is given by [tex] E = E_{0} (\frac{r}{R})^{3} [/tex]

    the assignment was to calculate Q of a piece of the cylinder of length L in two different ways, by gauss law in integral form and in differential form

    3. The attempt at a solution

    [tex]\oint E dA = \frac{Q}{e_{0}}[/tex]

    so [tex]Q = 2\pi e_{0} L E_{0} R[/tex]

    the other way around, [tex]\nabla E = \frac{\rho}{e_{0}}[/tex]

    so [tex] \rho = e_{0} E_{0} 3 \frac{r^{2}}{R^{3}} [/tex]

    so [tex] Q = e_{0} E_{0} 3 \oint \frac{r^{2}}{R^{3}} r dr d\theta dl = 2 \pi e_{0} E_{0} \frac{3}{4} L R [/tex]

    which is a factor 3/4 different from my other expression for Q

    where did I go wrong?
    Last edited: Feb 21, 2008
  2. jcsd
  3. Feb 21, 2008 #2


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    Homework Helper

    Review the form of the divergence in cylindrical coordinates. The non-vanishing term is (d/dr+1/r)E_r. That changes your '3' in rho to a '4'.
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