How to calculate the charge density in an electric field

[eschavez6]

I've been studying electric fields in class for some time and one thing is seemingly contradictory and really confuses me.

The charge density ρ is related to the electric field E and the permiativity ε and the potential \Phi by the following equation

ρ/ε=∇\cdotE=-∇²\Phi

if we examine the electric field created by a single point charge of magnitude q located at the origin, then the electrostatic potential can be expressed as follows

\Phi=\frac{q}{4πε(x^2+y^2+z^2)}

now I would expect the charge density in this system to be zero everywhere except the origin but if we take the laplacean of this electric field, instead we get

ρ/ε=-∇²\Phi=\frac{2q}{4πε(x^2+y^2+z^2)^2}

which is clearly non-zero.

is there an explanation for this discrepancy? Have I violated some fundamental assumption?

thanks in advance

 

[WannabeNewton]

Well you didn't really post your work so it's hard to tell where you went wrong. The Coulomb potential is much more elegantly expressed (and easier to work with) in spherical coordinates. $\varphi = \frac{Q}{4\pi \epsilon_0 }\frac{1}{r}$ hence $\nabla^{2}\varphi = \frac{Q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\partial_{r}(r^{2}\partial_{r}\frac{1}{r}) = 0$ for $r \neq 0$.

 

[nasu]

The potential goes like 1/r and not 1/r^2.
Try the correct form of the potential. It works even in Cartesian coordinates even though is more work.

 

[eschavez6]

thanks

it seems like i was using the wrong equation for electric potential. i had an r² in the denominator instead of just r