KleZMeR
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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.
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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.