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.