Integral Over all Space for Charge Density - Exponential Fun

[KleZMeR]

Homework Statement

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.
[/B]

Homework Equations

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

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.

 

[MisterX]

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.

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

 

[KleZMeR]

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

 

[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.