# A question about a charge density?

1. Nov 22, 2011

### wf198349

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

A static charge distribution produces radial electric field as follows:

$\stackrel{\rightarrow}{E}=A\frac{e^{-br}}{r}\stackrel{\rightarrow}{e_{r}}$

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

2. Relevant equations

$\nabla\cdot \stackrel{\rightarrow}{E}=\frac{\rho}{\epsilon}$

3. The attempt at a solution

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

But the result is wrong.Can you tell me why?

Last edited: Nov 22, 2011
2. Nov 22, 2011

### wf198349

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

A static charge distribution produces radial electric field as follows:

$\stackrel{\rightarrow}{E}=A\frac{e^{-br}}{r}\stackrel{\rightarrow}{e_{r}}$

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

2. Relevant equations

$\nabla\cdot \stackrel{\rightarrow}{E}=\frac{\rho}{\epsilon}$

3. The attempt at a solution

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

But the result is wrong.Can you tell me why?

Last edited: Nov 22, 2011
3. Nov 22, 2011

### vela

Staff Emeritus
What result did you get and what did you expect?

4. Nov 22, 2011

### wf198349

The reference answer is

$\rho=\epsilon_{0}\nabla\cdot \stackrel{\rightarrow}{E}=-\frac{\epsilon_{0}Ab}{r^{2}}e^{-br}+4\pi\epsilon_{0}A\delta(r)$

5. Nov 22, 2011

### vela

Staff Emeritus
It looks like there's a typo in the expression for the electric field. I think it should be
$$\mathbf{E} = \frac{Ae^{-br}}{r^2}\mathbf{e}_r$$

6. Nov 23, 2011

### wf198349

But how can I get the term $4\pi\epsilon_{0}A\delta(r)$ ?

7. Nov 23, 2011

### vela

Staff Emeritus
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.