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A potential function is given by $V(r)=\frac{Ae^{-\lambda r}}{r}$ Find charge density and hence charge.

I first took the gradient of potential to get $\vec{E}(r)=\frac{Ae^{-\lambda r}}{r}[\lambda+\frac{1}{r}] \hat{r}$

Now using Gauss's law $\nabla \cdot \vec{E}=\frac{\rho}{\epsilon_0} $
$\implies \rho=-\epsilon_0 \frac{Ae^{-\lambda r}}{r}[\frac{1}{r^2}+\frac{\lambda}{r}+\lambda^2]$

Now to find the charge $Q=\int \rho 4\pi r^2 dr$ .
Did I do it correctly? What limits should I choose for the integration?

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2 Answers 2

Gauss' law relates the integrated flux through a closed surface to the integrated charge inside it.

The symmetry of the situation lets you do the flux integral trivially, so I assume that you mean the volume one. Well, you integrate over the full angular ranges and from the center to the bounding surface.

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At least in the first $r$-derivation, you missed some $\lambda$'s. Generally, always check your units if you derive solutation like that, as for exmple "$1+\frac{1}{r}$" just must be flawed.

And you'll solve your problem more directly using the Poisson equation $\Delta V(r)=\frac{\varrho(r)}{\varepsilon_0}$, and that with a $\Delta$ in explicitly spherical coordinates.

To take a short track here, your $V(r)$ looks like a Yukawa potential, i.e. $$G(r)=\frac{\text{e}^{-\lambda r}}{4\pi\ r}$$ solves $(\Delta-\lambda^2)G(r)=-\delta(\vec r)$ and hence $$\Delta V(r)=4\pi A\left(-\delta(\vec r)+\frac{\lambda^2}{4\pi}\frac{\text{e}^{-\lambda r}}{r}\right)\equiv \frac{\varrho(r)}{\varepsilon_0}.$$

Integration over all of space, if you want to find the total charge. If you have a density like $\propto r^{-n}$, then clearly there are charges everywhere. Remark: For integration, notice that you're dealing with a three dimensional delta-function here (but from the Coulomb potential, you know that it gives you a single charge anyway). For the exp-integration, up to some small numbers, you can figure out the result just by power counting in $\lambda$'s.

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Sorry I missed out $\lambda$ in my equation I've now edited that –  Prakash Gautam Nov 9 '12 at 15:25
Sorry I am not much aware of delta function. Keeping $\delta(\vec r)$ in the result keeps the work of taking Laplacian still intact. So why should I prefer using $\delta(\vec r)$ in my answer of charge density? –  Prakash Gautam Nov 9 '12 at 15:30
@PrakashGautam: Not sure what you mean. Work of taking the Laplacian? I skipped taking derivatives, essentially, by "knowing" the Green function of a differential equation. The Delta-function is easy to integrate. In any case, have you used the gradient in Sperical cooridnates? –  NikolajK Nov 9 '12 at 18:17

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