# Integral Over all Space for Charge Density - Exponential Fun

1. Mar 12, 2015

### KleZMeR

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

The electric field is described by $E = \frac{C e^{-br}}{r^2}$,
find the charge density and then integrate over all-space and show it's zero.

2. Relevant equations

$E = \frac{C e^{-br}}{r^2} \\ \\$
$\nabla \cdot E = \frac{\rho}{\epsilon_0} \\ \\$
$\rho(r) = Q \delta^3 (r-r_0) \\ \\$

$I_{AllSpace} = \int_{0}^{\infty}\:f(r,\phi,\theta) \: r^2 \: dr \: sin\theta \: d\theta \: d\phi \\ \\$
$\nabla = \frac{d}{dr} \hat{r} \\ \\$
$\int_{0}^{\infty}f(r)Q \delta^3(r-r_0)dr = f(r_0) \\ \\$

3. The attempt at a solution

I tried to take $\nabla E = \frac{\rho}{\epsilon_0}$ and used the quotient solve for $\rho(r)$ and then integrated over all space and I do not get a zero.

With $\nabla \cdot E = (\frac{-e^{-ar}}{r} \:(ra+2))$
$\rho(r) = (\frac{-e^{-ar}}{r} \:(ra+2)) \: \epsilon_0$

and integrating over all space gives me

$4\pi \int_{0}^{\infty}\rho(r) \: r^2 \: dr$

and this gives me $(-1 + undefined) \: \epsilon_0$

my only other idea here is to use the delta function, which in the case of the radial integral does give me a zero due to the $r^2$ term being evaluated at zero.

$E_{AllSpace} = 4\pi \int_{0}^{\infty}\rho(r) \: r^2 \: dr = 4\pi Q \int_{0}^{\infty} \delta^3(r-r_0)r^2 \:dr = 0$

This second solution seems a bit short but I think it is still valid, I'm not sure if my first attempt if I made a mistake on the integral or if there is some other way of doing this? Any help would be appreciated.

2. Mar 12, 2015

### MisterX

What's the direction of the electric field vector? You stated it as if it's a scalar.

3. Mar 12, 2015

### KleZMeR

Ahh yes, it is in the $\hat{r}$ direction.

4. Mar 13, 2015

### KleZMeR

OK, so I found an issue and corrected for it, I was using $\nabla$ in cartesian coordinates, and it should have been in spherical.

My result after using spherical component for $r$ from $\nabla$, I get

$\rho(r)_{AllSpace} = 4\pi \int_{0}^{\infty} \frac{-bA}{\epsilon_0 } e^{-br} = A$

but I noticed that if I still use the delta function I get the desired answer of zero, but I think it is Q that I am trying to find? Any help is appreciated.

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