Proof of Gauss’ Theorem for a Rectangular Prism

Recall Gauss’ theorem, $\int\int\int_V (\vec{\nabla} \cdot \vec{F}) dV = \int\int_{S} (\vec{F} \cdot \vec{e}_{n}) dS$. This theorem can be written more precisely. The following statement of the Divergence theorem is a copy from reference [1]. Definitions are provided first. Volume $V$: Define $V$ as a region comprising three spatial dimensions. The volume has no holes in it. Boundary $\partial V$: Define $\partial V$ as a once-differentiable surface surrounding the volume $V$. The boundary has a thickness of zero and no holes in it. Points not in $V$ are not in $\partial V,$ […]

Overview of Gauss’ Theorem and Basics of Integration

It turns out that deriving Gauss’ Law is easier said than done. There are several steps according to a StackExchange post [1]. The first of these steps is understanding Gauss’ Theorem. Hmm. Perhaps Gauss used his own theorem to derive his electrostatics law. After a quick online search, it is clear that Gauss’ Theorem is just another name for the Divergence theorem [2]. The Divergence Theorem is [3] $\int\int\int_V (\vec{\nabla} \cdot \vec{F}) dV = \int\int_{S} (\vec{F} \cdot \vec{e}_{n}) dS$. Oof. That is a lot of symbolism to break down. Fortunately I am able to break this down;

Gauss’ Law, Part 1

In Derivation #3, the expression, $\frac{\partial \beta(t)}{\partial t}$, was written. This is an expression for a derivative of a function $\beta(t)$. Now that a derivative has been introduced, Maxwell’s equations can be investigated. I start with Gauss’ Law. But first, slightly more information about derivatives is needed. I can consider the pieces of $\frac{\partial \beta(t)}{\partial t}$ separately. These pieces are $\beta(t)$ and $\frac{\partial}{\partial t}$. The $\beta(t)$ is, of course, a function while $\frac{\partial}{\partial t}$ is an operator called a differential operator that forms $\frac{\partial \beta(t)}{\partial t}$ if $\frac{\partial}{\partial t}$ is