Potential generated by a point charge in a isotropic medium

[gabu]

Homework Statement

When a point charge is positioned at the origin = 0 in an isotropic
material, a separation of charge occurs around it, the Coulomb field of the
point charge is screened, and the electrostatic potential takes the form

\phi(r) = \frac{A}{r} \exp\left( -\frac{r}{\lambda} \right)

Here, r is the distance from the origin, and A and are constants. In this case,
determine the charge density distribution () in the space surrounding the
origin.

(b)
In formula (1) above, the potential becomes infinitely large at the origin. This
shows that the point charge is located at the origin. Using Gauss’s Law, find
the charge of the point particle.

Homework Equations

[/B]
Electric Potential
\phi(r) = \frac{A}{r} \exp\left( -\frac{r}{\lambda} \right)

Laplacian in spherical coordinates
\Delta = \frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2\,\frac{\partial}{\partial r}\right)

The Attempt at a Solution

To solve the first part, I have used Poisson's equation \Delta \phi(r) = -\frac{\rho}{\epsilon_{0}}

and obtained

\rho = \frac{A\epsilon_{0}}{r\,\lambda^2}\,\exp\left( -\frac{r}{\lambda}\right)

for the charge distribution surrounding the origin. For the second part of the problem I used the fact that the potential is spherically symmetrical to write Gauss' law as

\vec{\nabla}\phi\cdot \hat{r}\int dA = \frac{Q}{\epsilon_{0}}
\vec{\nabla}\phi\cdot \hat{r}(4\pi\epsilon_{0}) = \frac{Q}{\epsilon_{0}}
finally obtaining

q = -4\pi\epsilon_{0}A\left(1-\frac{r^2}{\lambda}\right) \,\exp\left( -\frac{r}{\lambda}\right)

My problem with this solution is that it depends on the distance from the origin, but the problem said it was a point charge. The only thing I can think about to go around this is considering r=0, but I don't have a good justification for this. What did I get wrong?

Thank you very much.

 

[Orodruin]

You are computing the charge within a radius $r$. This includes both the point charge at the origin and any charge from the continuous distribution you computed in (a) up to the radius $r$. In order to get the point charge only, you need to take the limit $r\to 0$.