This is said to be the equation to find the electric field for a continuous distribution of charges:
Just starting to learn calc III, I'm not exactly sure how to interpret this. I've never seen the prime notation paired with a vector like this before. What, geometrically does each symbol represent? I understand well electromagnetism and how Coulomb's Law and Gauss' Law apply intuitively, but the fact that the unit vector is a function multiplied by the charge density of a function, I'm really confused by. Especially because I thought $\rho$ was a function of $(x,y,z)$. I'm trying to get a grasp of the idea behind this rather than a necessarily mathematically rigorous definition of what's happening.
The primed coordinates are just "dummy" coordinates over which we integrate the contributions to the electric field at a fixed observation point $\vec r$ from a differential charge, $dq=\rho(\vec r')\,dV'$, at $\vec r'$.
The distance between to fixed observation point $\vec r$ and the differential charge at $\vec r'$ is $|\vec r-\vec r'|$ and the unit vector from the charge to the observation point is $\frac{\vec r-\vec r'}{|\vec r-\vec r'|}$.
The contribution to the electric field, $d\vec E(\vec r)$, from a differential charge $dq=\rho(\vec r')\,dV'$, at $\vec r'$ is given by
$$\begin{align} d\vec E(\vec r)&=\frac{dq}{4\pi \epsilon_0 |\vec r-\vec r'|^2}\,\,\underbrace{\left(\frac{\vec r-\vec r'}{|\vec r-\vec r'|}\right)}_{=\hat u(\vec r-\vec r')}\\\\ &=\frac{\rho(\vec r')\,\hat u(\vec r-\vec r')}{4\pi \epsilon_0 |\vec r-\vec r'|^2}\,dV' \end{align}$$
Then we sum the contributions by integrating over the volume $V$ that contains all of the charge to obtain'
$$\begin{align} \vec E(\vec r)&=\iiint_V d\vec E(\vec r)\\\\ &=\iiint_V\frac{\rho(\vec r')\,\hat u(\vec r-\vec r')}{4\pi \epsilon_0 |\vec r-\vec r'|^2}\,dV' \end{align}$$