Confusion about Gauss's law for Electrostatics

Question

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?

Answer 1

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

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

Answer 2

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.