Why is the Charge Density Calculation Incorrect?

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Homework Help Overview

The discussion revolves around the calculation of charge density from a given radial electric field expression. The electric field is defined as \(\stackrel{\rightarrow}{E}=A\frac{e^{-br}}{r}\stackrel{\rightarrow}{e_{r}}\), where \(A\) and \(b\) are constants. Participants are attempting to compute the charge density using the divergence of the electric field.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are examining the divergence of the electric field and questioning the correctness of their results. There is discussion about potential typos in the electric field expression and how to derive specific terms in the charge density calculation.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the electric field expression and its implications for charge density. Some guidance has been offered regarding the application of Gauss's law, but no consensus has been reached on the correct approach or final result.

Contextual Notes

There are indications of potential typos in the electric field expression, and participants are considering the implications of these errors on their calculations. The presence of a delta function term in the charge density is also under scrutiny.

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