Interpretation of Gauss's theorem applied to Maxwell's equations: $\dfrac{d}{dt} \int \rho \ dV + \int \mathbf{j} \cdot \mathbf{n} \ dS = 0$ Using Maxwell's equations and Gauss's theorem, we get
$$\dfrac{d}{dt} \int \rho \ dV + \int \mathbf{j} \cdot \mathbf{n} \ dS = 0,$$
where $\rho$ is the electric charge density and $\mathbf{j}$ is the electric current density.
Therefore, we have that
$$e = \int \rho \ dV$$
is the total charge.
Electric current is said to be $I = \dfrac{Q}{t}$, where $Q$ is quantity of charge in coulombs. So does this mean that the term $\dfrac{d}{dt} \int \rho \ dV$ is the current? After all, this is the change in total electric charge with respect to time, which matches $I$, right?
Or does this mean that $\int \mathbf{j} \cdot \mathbf{n} \ dS$ is the electric current?
What do each of these two terms actually represent?
I would greatly appreciate it if people would please take the time to clarify this.
 A: $\int \mathbf{j} \cdot \hat{\mathbf{n}} \, \mathrm{d} S$ is the integral of current density over a surface; this is equal to the total current passing through this surface. In your notation you could call this $I$.
$\int \rho \, \mathrm{d} V$ is the integral of charge density over a volume; this is equal to the total charge in the volume. You can denote this $Q$.
So the statement of Gauss's law can be written
$$I = -\frac{\mathrm{d} Q}{\mathrm{d} t};$$
the minus sign comes from the orientation of the surface: we define current flowing out of the surface to have a positive sign.
A: The continuity equation states that the charge of a volume V can only change due to the electric flow (i.e. when they move outside or inside the volume). The same continuity equation holds for liquids as well, with $\varrho$ being the density and $j=\varrho v$, where $v$ is the velocity of the flow.
These continuity equations does not have source, but a general one do have it. For example, the continuity equation for the energy of the electromagnetic field is
$$\frac{\partial w}{\partial t}+\text{div}S=-jE$$
Where $w$ is the energy density, i.e.
$$w=\frac{1}{2}(ED+BH)$$
And $S$ is the Poynting-vector:
$$S=E\times H$$
Integrating this equation to the whole space, we get that the total  energy can only change due to the external power $jE$.
