# A question about a charge density?

1. Homework Statement [/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.

## Homework Equations

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

## 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:

1. Homework Statement [/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.

## Homework Equations

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

## 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:
vela
Staff Emeritus
Homework Helper
What result did you get and what did you expect?

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

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

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

vela
Staff Emeritus