# Charge density for a given potential

1. Feb 18, 2017

### Silviu

1. The problem statement, all variables and given/known data
Given the electric potential $V(r)=A\frac{e^{-\lambda r}}{r}$ calculate the charge density $\rho(r)$ and the electric field $E(r)$.
They specify the answer for charge density should be: $\rho = \epsilon_0 A(4\pi \delta^3(r)-\lambda^2e^{-\lambda r}/r)$

2. Relevant equations
$E=-\nabla V$
$\nabla E = \frac{\rho}{\epsilon_o}$

3. The attempt at a solution
For the electric field I got: $E = Ae^{-\lambda r}(1+\lambda r)/r^2$, pointing in the r direction and this is the same as their answer. Then I tried to take the divergence of E, and in spherical coordinates this would be $\nabla E = \frac{1}{r^2}\frac{\partial r^2 E}{\partial r} = \frac{1}{r^2}\frac{\partial (Ae^{-\lambda r}(1+\lambda r))}{\partial r}$ . However this doesn't give me a delta function so I am not sure where I did something wrong.

2. Feb 18, 2017

### vela

Staff Emeritus
Your expression for the divergence is only valid for $r\ne 0$. You need to consider what's happening at $r=0$ in another way.

3. Feb 19, 2017

### Silviu

Thank you! So I found a solution online to this problem but I am still a bit confused. I attached their solution. So i understand how they obtain the delta function but when they apply the divergence to the other term, they use $\frac{\partial}{\partial r}$. Shouldn't they use the divergence in spherical coordinates which is $\frac{1}{r^2}\frac{\partial (r^2 E')}{\partial r}$, with E' being the term that remained to be acted on?

#### Attached Files:

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4. Feb 19, 2017

### TSny

They are using the identity

where $\psi$ is a scalar function.

(See https://en.wikipedia.org/wiki/Vector_calculus_identities#Product_of_a_scalar_and_a_vector)

The second term on the right does not involve a divergence. It involves a gradient.