# Confusion about Gauss's law for Electrostatics

I just learning about Gauss's law in integral and differential form. There's something I'm a bit confused about:

Let $\vec{r}$ be the location of your test charge with respect to the origin, and $\vec{r'}$ be the location of a charge $dq$ on a charge distribution. Gauss's law says:

$$\nabla\cdot\vec{E}(\vec{r})=\frac{\rho(\vec{r'})}{\epsilon_0},$$ and as a corollary, $$\nabla^2V(\vec{r})=-\frac{\rho(\vec{r'})}{\epsilon_0}.$$ I'm a bit confused because on the left hand side of each of these equations, we get a function of $\vec{r}$, so we can evaluate it a any point in the space, but on the right hand side, we get a function of $\vec{r'}$, so the right hand side is evaluated at points on the charge distribution. What am I not understanding?

Typically, one defines a variable with respect to some observation point, P. In this case, the $\vec{r}$ is a vector pointing from the origin (defined by your chosen frame of reference and coordinate basis) to point P (which happens to be the location of your point charge $q$). The $\vec{r}$' here would be a vector from point P to the source of the field (i.e., the charge distribution $dq$).

Remember, you observe a field at a given point in space with respect to a source at some other location. Generally one determines the field or potential for arbitrary observation points and arbitrary charge distributions.

You're getting confused because the author you're reading has expressed Gauss' law using unusual and confusing notation, to the point where I would actually call it incorrect.

I can sort of see how the confusing equations could arise from a derivation of Gauss' law from Coulomb's law. From Coulomb's law, you have

$$\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(\vec{r}')(\vec{r}-\vec{r}')}{|\vec{r}-\vec{r}'|^3} \, d^3 \vec{r}'\ .$$

Taking the divergence of both sides with respect to $\vec{r}$, and taking advantage of the identity

$$\nabla \cdot \left(\frac{\vec{r}}{|\vec{r}|^3}\right) = 4\pi \delta(\vec{r})$$

along the way, you end up with

$$\nabla\cdot\vec{E}(\vec{r}) = \frac{1}{\epsilon_0} \int \rho(\vec{r}')\ \delta(\vec{r}-\vec{r}')\, d^3 \vec{r}'\ .$$

But this is where I disagree with how your author has expressed things. $\delta(\vec{r}-\vec{r}')$ is zero everywhere except where $\vec{r}=\vec{r}'$. The normal way to express $\delta$'s sifting property on that last integral would be to say that the integral evaluates to $\rho(\vec{r})$, not to say that the integral evaluates to $\rho(\vec{r}')$. If you say that the integral evaluates to $\rho(\vec{r}')$, you'd have to add that the only $\vec{r}'$ involved is where $\vec{r}'=\vec{r}$, in which case it make more sense to just say $\rho(\vec{r})$ instead of $\rho(\vec{r}')$. And that would indeed be what is normally done; Gauss' law would more commonly be expressed as

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

• Thank you so much. :) This really cleared things up for me. Sep 19 '14 at 19:45