# Singularity of potential

1. Oct 19, 2016

### spaghetti3451

1. The problem statement, all variables and given/known data

Determine the electric field and the charge density associated with the potential $\displaystyle{V(r)=A\frac{\exp{-\lambda r}}{r}}$, where $A$ and $\lambda$ are constants.

2. Relevant equations

3. The attempt at a solution

The electric field is easy to determine:

$\displaystyle{{\bf{E}} = -\nabla V}$
$\displaystyle{= -A\ \nabla\left(\frac{\exp\left(-\lambda r\right)}{r}\right)}$
$\displaystyle{= -A\ \left(\frac{\partial}{\partial r}\left(\frac{\exp\left(-\lambda r\right)}{r}\right),0,0\right)}$
$\displaystyle{= A\ \left(\frac{(1+\lambda r)\exp\left(-\lambda r\right)}{r},0,0\right)}$

But the electric field is singular at the origin.

How does this affect the calculation of the charge density?

2. Oct 19, 2016

### Fightfish

Well, that means you have a point charge at the origin - recall that $\nabla \cdot \left(\frac{\vec{r}}{r^2}\right) = 4 \pi \delta(r)$.

3. Oct 19, 2016

### spaghetti3451

In which step have I made the mistake?

$\rho = \epsilon_{0}\ \nabla\cdot{{\bf{E}}}$

$= \epsilon_{0}\ A\ \nabla\cdot{\left((1+\lambda r)\frac{\exp\left(-\lambda r\right)}{r},0,0\right)}$

$= \epsilon_{0}\ \frac{A}{r^{2}}\ \frac{\partial}{\partial r} \left(r^{2}(1+\lambda r)\left(\frac{\exp\left(-\lambda r\right)}{r}\right)\right)$

$= \epsilon_{0}\ \frac{A}{r^{2}}\ \frac{\partial}{\partial r} \left[(r+\lambda r^{2})\exp\left(-\lambda r\right)\right]$

$= \epsilon_{0}\ \frac{A}{r^{2}}\ \left[(1+2\lambda r)\exp\left(-\lambda r\right)-\lambda(r+\lambda r^{2})\exp\left(-\lambda r\right)\right]$

$= \epsilon_{0}\ \frac{A}{r^{2}}\ \left[(1+2\lambda r)-\lambda(r+\lambda r^{2})\right]\exp\left(-\lambda r\right)$

$= \epsilon_{0}\ A\ \left[1+\lambda r-(\lambda r)^{2})\right]\frac{\exp\left(-\lambda r\right)}{r^{2}}$

4. Oct 19, 2016

### Fightfish

Here:

5. Oct 20, 2016

### spaghetti3451

No, the signs are right; its the denominator that should be $r^2$, no?