0
$\begingroup$

I have questions regarding the normal vector that we use in calculating electric flux and Gauss's law.

  1. In the first and second figures, why is the normal vector is in that direction? and why not to the other side? enter image description here
  1. Why do when we consider a 3D object we always choose the normal vector pointing outside?
$\endgroup$
5
  • $\begingroup$ Because that's the way that Gauss' law works. The normal is always the outward pointing one. If you want to use the inward pointing normal then you need to change the sign of the integral. $\endgroup$
    – mike stone
    Aug 9 at 11:37
  • $\begingroup$ @mikestone How about the first question? Why do we pick the normal vector like the figures and not be directed to the other side ? Do we just choose the normal vector that "follows" the direction of the electric field? $\endgroup$ Aug 9 at 11:57
  • $\begingroup$ the direction of normal is not important, what matters is the direction of electric field. It is a bit like when you draw xyz coordinates. Does it matter which way they point? No. But once you define the direction, you can talk about which way your (e.g.) particle is moving $\endgroup$
    – Cryo
    Aug 9 at 12:57
  • $\begingroup$ Related : How to know the direction of the unit normal vector to an open surface?. $\endgroup$
    – Frobenius
    Aug 9 at 14:08
  • $\begingroup$ Ah I see now, thank you for your responses. $\endgroup$ Aug 10 at 4:16
3
$\begingroup$

It should be emphasized that the concept of flux depends on three things

  • A field $\mathbf{F}$ whose "flux" we're trying to calculate (eg electric field, magnetic field, gravitational field whatever)
  • A surface $S$ across which we're trying to calculate the flux
  • A choice of orientation for the surface $S$ (for a $2$-dimensional surface in a $3$ dimensional space, it is enough to specify a continuously varying normal vector field $\mathbf{n}$)

Thus, the proper terminology is "the flux of the vector field $\mathbf{F}$, across the surface $S$ with respect to the normal vector field $\mathbf{n}$", and the definition for this is an integral: \begin{align} \Phi_{\mathbf{F},S,\mathbf{n}}:=\int_S\mathbf{F}\cdot \mathbf{n}\,dA. \end{align}

Usually, people aren't so explicit with terminology, and may simply write "the flux of $\mathbf{F}$ across $S$", or simply "the flux of $\mathbf{F}$".


In the case where the surface "encloses a volume" $\Omega$, such as the surface of a sphere, which is the boundary of a ball, we always choose $\mathbf{n}$ to "point outwards" because then, the divergence theorem states \begin{align} \int_{\Omega}\text{div}(\mathbf{F})\,dV&=\int_{\partial \Omega}\mathbf{F}\cdot \mathbf{n}\,dA \end{align} So, if we instead considered the inward pointing normal vector field $\mathbf{n}_{in}:=-\mathbf{n}$, then the divergence theorem must be modified to \begin{align} \int_{\Omega}\text{div}(\mathbf{F})\,dV&=-\int_{\partial \Omega}\mathbf{F}\cdot \mathbf{n}_{in}\,dA. \end{align} We don't like carrying around excess minus signs in our formulas, that's why we always choose the outward pointing normal. Note that this is a choice, not a forced requirement. I could (if I was a masochist) use the inward-pointing normal and simply carry around that minus sign everywhere I go.


In the case of a surface which doesn't enclose any region, such as a planar region like in the figure, there is no notion of inside/outside, and there is no other reasons which indicate one choice of sign is to be preferred over another (in the above case, the divergence theorem atleast suggests to us that outward normal makes the formulas look nicer).

Hence, in your picture, one has to declare by-hand that we orient the planes using the normal vector field $\mathbf{n}$ pointing in a certain direction. You're completely free to choose which, but once you make a choice, stick with it, and do not change your choice in the middle of a calculation (unless you want to confuse yourself and everyone else :). Having made this choice, one can then unambiguously speak of "the electric flux across the plane with respect to the normal $\mathbf{n}$".

$\endgroup$
2
  • 1
    $\begingroup$ This answer is excellent. It may also be worth noting that a choice of orientation for an open surface induces a natural orientation on its boundary, which is why the ambiguity in the choice of $\mathbf n$ is not problematic for the integral versions of the Faraday and Ampere laws. $\endgroup$
    – J. Murray
    Aug 11 at 13:26
  • 1
    $\begingroup$ @J.Murray thank you, and indeed that's a very good remark, and just elaborating slightly for future readers who may not already be aware : Given a surface $S$ oriented by a normal $\mathbf{n}$, the boundary $\partial S$ will have two possible orientations. One of these however is the "natural induced one", and it is "natural" because then Stokes' formula reads $\int_S\text{curl}(\mathbf{F})\cdot \mathbf{n}\,dA = \int_{\partial S}\mathbf{F}\cdot \mathbf{t}\, dl$ ($\mathbf{t}$ is the 'positively-oriented' unit tangent to $\partial S$,), and there are no rogue minus signs floating around. $\endgroup$
    – peek-a-boo
    Aug 11 at 13:44
0
$\begingroup$

This is how Gauss's law was formulated.

For an open surface: $$\phi_{E} = \int \vec E.d \vec A$$ where the angle $\theta$ between $\vec E$ and $d \vec A$ is chosen such that $\theta \leq 90^{\circ}$

For a closed surface: $$\phi_{E} = \oint \vec E.d \vec A$$ where the direction of $d \vec A$ is chosen such that $d \vec A$ points normally outwards from the closed Gaussian surface.

(All the symbols have their usual meaning)

Hope this helps.

$\endgroup$
3
  • 1
    $\begingroup$ for open surface your definition will lead to inconsistencies with fields that change orientation or are inhomogeneous on the scale of the surface. IMHO it is better to fix normal direction as an explicit arbitrary choice $\endgroup$
    – Cryo
    Aug 9 at 13:01
  • $\begingroup$ Yes it isn't extensive, but I believe it is appropriate for the level of understanding the author of the question is at currently. Similar defintions are provided in several freshman introductory texts too. $\endgroup$
    – Cross
    Aug 9 at 15:08
  • $\begingroup$ Okay, thank you for your answer. $\endgroup$ Aug 10 at 4:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.