# How to find the charge density given by a Tricky Potential?

## Homework Statement

I am trying to solve Problem 2.45 in Electrodynamics by Griffiths, however, my answers were different from those in the book, I am suspect I got a missing step but I could not find it, so here is the Given Problem

Find the charge density $\rho$ given by a potential $$V=A\frac{e^{-\lambda r}}{r}$$ where $A$ and $\lambda$ are constant.

## Homework Equations

$$E=-\nabla V$$ $$\nabla\cdot E = \frac{\rho}{\varepsilon_0}$$

## The Attempt at a Solution

So what I did is, first, find for the E-field using: $$E=-\nabla V,$$ so $$E=-\hat{r}\frac{\partial }{\partial r} A\frac{e^{-\lambda r}}{r}$$ by employing chain rule, the E-field is now given by $$E= A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r}.$$

Now, the charge density $\rho$ can be obtained thru $$\rho=\varepsilon_0\nabla\cdot E,$$ substituting the E-field to the equation, we will get $$\rho=\varepsilon_0\nabla\cdot \left [A\lambda\frac{ e^{-\lambda r}}{r}\hat{r}+A\frac{ e^{-\lambda r}}{r^2}\hat{r} \right ],$$ evaluating again using chain rule, we will get $$\rho=\varepsilon_0 \left [A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r} +\frac{1}{r}\left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A\lambda e^{-\lambda r} \right ) + A\lambda e^{-\lambda r}\nabla\cdot\frac{\hat{r}}{r^2} +\frac{1}{r^2} \left( \frac{1}{r^2}\frac{\partial }{\partial r} r^2 A e^{-\lambda r} \right ) \right ],$$ this will eventually become a bit more messy but the charge density will become $$\rho=A \varepsilon_0 e^{-\lambda r} [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2}{r} ]$$ however, this a bit different from Griffiths' answer of $$\rho=A \varepsilon_0 [ 4\pi \delta^{3}(\mathbf{r})-\frac{\lambda^2 e^{-\lambda r}}{r}]$$

where did I went wrong?

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