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A question about a charge density?

  • Thread starter wf198349
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  • #1
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1. Homework Statement [/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.

Homework Equations



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

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:

Answers and Replies

  • #2
4
0
1. Homework Statement [/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.

Homework Equations



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

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:
  • #3
vela
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What result did you get and what did you expect?
 
  • #4
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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]
 
  • #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]
 
  • #6
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But how can I get the term [itex]4\pi\epsilon_{0}A\delta(r)[/itex] ?
 
  • #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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