# Divergence of E

1. Nov 15, 2009

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

The electrostatic field of a point charge q is E=$$\frac{q}{4 \pi \epsilon r^3}$$ r. Calculate the divergence of E. What happens at the origin?

2. Relevant equations

3. The attempt at a solution

Well the solution is: $$\nabla$$.E= $$\partial$$Ex$$/$$$$\partial$$x + $$\partial$$Ey$$/$$$$\partial$$y + $$\partial$$Ez$$/$$$$\partial$$z

Ex= $$\frac{qx}{4 \pi \epsilon r^3}$$ and Ey=$$\frac{qy}{4 \pi \epsilon r^3}$$ and Ez=$$\frac{qz}{4 \pi \epsilon r^3}$$ and r= $$\sqrt{x^2 + y^2 + z^2}$$

After calculation I found the result $$\nabla$$.E= $$\partial$$Ex$$/$$$$\partial$$x + $$\partial$$Ey$$/$$$$\partial$$y + $$\partial$$Ez$$/$$$$\partial$$z= 0
Is it correct? I think it is wrong! Then why it is wrong?
Somewhere else I saw that the result was $$\nabla$$.E= $$\frac{\rho}{\epsilon}$$ !

2. Nov 16, 2009

### clamtrox

What is $$\rho$$ in this case (a point particle)? It's definitely zero at most places, but where is it not?

3. Nov 16, 2009

I really don't know what $$\rho$$ is here! But do you mean that I calculated it in the correct way?

4. Nov 16, 2009

### clamtrox

All the charge of a pointlike particle is located at a single point (that's why it's called pointlike). At what coordinates is the charge located in this case? What values does the electric field get there?

5. Nov 16, 2009

Oh thank you for telling me the definition of a pointlike particle, because I didn't know that all the charge is located at a single point! The coordinate is cartesian. I have to translate the main problem, well I will try.
The main problem is this:

If E is the electrostatic field and $$\Phi$$ is the potential of the electrostatic field prove that: $$\int$$$$\rho$$$$\Phi$$dv= $$\epsilon$$$$\int$$ E2dv by using this theorem
$$\nabla$$.($$\Phi$$E)=$$\Phi$$($$\nabla$$.E)+($$\nabla$$$$\Phi$$).E

It is actually an example and in the solution it says that we know $$\nabla$$.E=$$\frac{\rho}{\epsilon}$$ and also E=-$$\nabla$$$$\Phi$$ and also it says that $$\int$$$$\nabla$$.($$\Phi$$E)dv= 0
Do you think it's a correct problem? I mean is it right or wrong, the solution!

6. Nov 16, 2009

### clamtrox

So do you now know what the charge distribution is?

You also misunderstood my question. At what location is the particle in? In terms of x,y,z? And what is the value of the electric field there?

This problem has nothing to do with the question you asked earlier :) To solve it, you only have to use the formulas that you are given.