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A question about a charge density?

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data[/b]

    A static charge distribution produces radial electric field as follows:

    [itex]\stackrel{\rightarrow}{E}=A\frac{e^{-br}}{r}\stackrel{\rightarrow}{e_{r}}[/itex]

    A,b are constants,please compute the charge density.

    2. Relevant equations

    [itex]\nabla\cdot \stackrel{\rightarrow}{E}=\frac{\rho}{\epsilon}[/itex]

    3. The attempt at a solution

    [itex]\nabla\cdot \stackrel{\rightarrow}{E}=\frac{1}{r^{2}}\frac{ \partial}{\partial r}(r^{2}A\frac{e^{-br}}{r})=\frac{1}{r^{2}}\frac{\partial}{\partial r}(Are^{-br})[/itex]

    But the result is wrong.Can you tell me why?
     
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2
    1. The problem statement, all variables and given/known data[/b]

    A static charge distribution produces radial electric field as follows:

    [itex]\stackrel{\rightarrow}{E}=A\frac{e^{-br}}{r}\stackrel{\rightarrow}{e_{r}}[/itex]

    A,b are constants,please compute the charge density.

    2. Relevant equations

    [itex]\nabla\cdot \stackrel{\rightarrow}{E}=\frac{\rho}{\epsilon}[/itex]

    3. The attempt at a solution

    [itex]\nabla\cdot \stackrel{\rightarrow}{E}=\frac{1}{r^{2}}\frac{ \partial}{\partial r}(r^{2}A\frac{e^{-br}}{r})=\frac{1}{r^{2}}\frac{\partial}{\partial r}(Are^{-br})[/itex]

    But the result is wrong.Can you tell me why?
     
    Last edited: Nov 22, 2011
  4. Nov 22, 2011 #3

    vela

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    What result did you get and what did you expect?
     
  5. Nov 22, 2011 #4
    The reference answer is

    [itex]\rho=\epsilon_{0}\nabla\cdot \stackrel{\rightarrow}{E}=-\frac{\epsilon_{0}Ab}{r^{2}}e^{-br}+4\pi\epsilon_{0}A\delta(r)[/itex]
     
  6. Nov 22, 2011 #5

    vela

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    It looks like there's a typo in the expression for the electric field. I think it should be
    [tex]\mathbf{E} = \frac{Ae^{-br}}{r^2}\mathbf{e}_r[/tex]
     
  7. Nov 23, 2011 #6
    But how can I get the term [itex]4\pi\epsilon_{0}A\delta(r)[/itex] ?
     
  8. Nov 23, 2011 #7

    vela

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    Apply the integral form of Gauss's law using a sphere of radius r as the Gaussian surface and take the limit as r goes to 0. From that result, you should be able to deduce a point charge has to reside at the origin.
     
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