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.

  • 1
    $\begingroup$ The integral represents the flux of $\vec{E}$ across the surface $S$ and you can decompose $\vec{E}$ into any coordinates you like but the components not parallel to $\hat{n}$ at a point on the surface don't cross the surface at that point, so there is no contribution to the flux from those components. So the flux is only contributed by those components parallel to $\vec{n}$. $\endgroup$ – AEngineer Jan 24 '20 at 4:57
  • $\begingroup$ @AEngineer Oh, that makes complete sense. Thanks for the clarification! $\endgroup$ – The Pointer Jan 24 '20 at 4:58

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.

  • $\begingroup$ Ahh, yes, that makes sense. Thanks for the clarification! $\endgroup$ – The Pointer Jan 24 '20 at 5:10

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