# Confusion about Gauss' law differential form

1. Jan 2, 2012

### constfang

http://einstein1.byu.edu/~masong/emsite/S1Q50/EQMakerSL1.gif

Hi guys, I have a little confusion about the Gauss' law in differential form over here, obviously, many textbook wrote it in the above form, but actually, the only place at which divE is not zero is at the locations where the charges are present. so I read over here: http://farside.ph.utexas.edu/teaching/em/lectures/node30.html#e343 (equation 208) that there should be a Dirac delta function over there, is it correct? then why does people keep omitting that Dirac delta function and just quoted it as above, where no information regarding the position of the point where we're calculating the divergence is given? Thank you.

2. Jan 2, 2012

### vanhees71

The fundamental law, using SI units, is indeed

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

Here, $\rho$ is the charge density of the charged matter, producing the electromagnetic field.

If the matter consists of a single charge at rest, this density is a $\delta$ distribution

$$\rho_{\text{single charge}}(\vec{x})=q \delta^{(3)}(\vec{x}-\vec{x}_0).$$

If you have condensed matter, you can coarse grain the sum of the single-charge [itex]\delta[itex] distributions to a continuous charge distribution.

3. Jan 2, 2012

### constfang

Oh I got it now, thank you very much.