- #1

Marcus95

- 50

- 2

## Homework Statement

A static charge distribution has a radial electric field of magnitude

##E = \alpha \frac{e^{-\lambda r}}{r} ##

where λ and α are positive constants. Calculate the total charge of the distribution.

## Homework Equations

Gauss's law ##Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S}##

##\rho/\epsilon_0 = \nabla \cdot \vec{E} ##

## The Attempt at a Solution

I have tried two ways to go about this problem, without success. First I tried using Gauss's law, placing a Gaussian sphere of radius a on the distribution:

##Q/\epsilon_0 = \int \vec{E} \cdot d\vec{S} = 4\pi\alpha (ae^{-\lambda a}) ##

We should now find the total charge by letting ## a \rightarrow \infty ##, but ##\lim_{a \rightarrow \infty} ae^{-\lambda a} = 0##, giving ##Q = 0## which i thought cannot be right.*

My second method involved finding the charge distribution

##\rho/\epsilon_0 = \nabla \cdot \vec{E} = \alpha \frac{e^{-\lambda r}(1-\lambda r)}{r^2} ## but this gives an integral which is divergent!

*I have done some further thinking and realized that ##Q_{tot} = 0## might be the answer, because the charge density changes sign at ##r = 1/\lambda## so the total net charge could as well be zero.

However, how is it possible that the volume integral of #\rho# is divergent then? Or is it not?