Integral form of Gauss's law: dot product demands that we find the part of $\vec{E}$ parallel to $\hat{\mathbf{n}}$ (perpendicular to the surface)?

Question

I am currently studying the textbook A Student's Guide to Maxwell's Equations by Daniel Fleisch. In a section discussing the integral form of Gauss's law, the author says that the dot product in

$$\oint_S \vec{E} \cdot \hat{\mathbf{n}} \ da = \dfrac{q_{enc}}{\epsilon_0}$$

"tells you to find the part of $\vec{E}$ parallel to $\hat{\mathbf{n}}$ (perpendicular to the surface)".

Why does the presence of the dot product demand that we find the part of $\vec{E}$ parallel to $\hat{\mathbf{n}}$ (perpendicular to the surface)? The only degenerate case seems to be when $\vec{E}$ is perpendicular to $\hat{\mathbf{n}}$, which means that their dot product is equal to zero. Besides this case, I do not see why $\vec{E}$ must be precisely parallel to $\hat{\mathbf{n}}$?

I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

The electric flux is defined as the sum of the components of the field that are parallel to the normal of the surface. The dot product gives you exactly that: given two vectors, say $\mathbf E$ and $\hat{\mathbf n}$, their dot product is given as

$$\mathbf E \cdot \hat{\mathbf n} = \lvert \mathbf E \rvert\lvert\hat{\mathbf n}\rvert\cos\theta$$

The projection of $\mathbf{E}$ onto $\hat{\mathbf{n}}$ is given by

$$ \mathbf{E}_n = \lvert \mathbf E \rvert\cos\theta $$

however since $\lvert \hat{\mathbf{n}}\rvert = 1$, this is equivalent to taking the dot product.