# Divergence of Point Charge

1. Oct 16, 2009

### hbweb500

So I am playing around with the differential form of Gauss's Law:

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$

Starting off simple with a point charge, the electric field is:

$$\vec{E} = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \hat{r}$$

And the divergence, in spherical coordinates, is:

$$\nabla \cdot \vec{E} = \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 E_r)$$

$$= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \right)$$

$$= \frac{1}{r^2} \frac{\partial}{\partial r} \left( \frac{q}{4 \pi \epsilon_0}} \right)$$

$$= 0$$

I can handle this much. It makes sense that the divergence is everywhere 0, since the only charge density is a point charge. Griffiths has a discussion about this very thing, where he states that the infinite divergence at the origin causes things to work out as expected.

My problem is in dealing with a cartesian coordinate system. I didn't recall the divergence equation in spherical coordinates when I was first playing around with this, so I tried it in cartesian. Here:

$$\vec{E} = \frac{q}{4\pi \epsilon_0} \frac{ x \hat{x} + y \hat{y} }{(x^2+y^2)^{\frac{3}{2}}}$$

But the divergence of this is proportional to:

$$\nabla \cdot \vec{E} \propto \frac{-2}{(x^2 + y^2)^3}$$

Which clearly isn't zero everywhere. I've checked my divergence and electric field equation, but I can't find the difference between it and the spherical ones I am using.

So, what gives? Is the divergence in spherical coordinates 0, but nonzero in cartesian?

2. Oct 16, 2009

### Mapes

Isn't there a z axis in Cartesian coordinates? Add this and double check your divergence calculations.

3. Oct 17, 2009

### hbweb500

Oh, yes, that would be it. I suppose the equations don't like it very much when you try to restrict the electric field from a point charge to a plane...

Thanks!