I've been trying to wrap my head around Walter Lewin's lecture on Gauss's law and electric flux and I can't go on without thoroughly getting this first. I think I've understood the electric flux part, but then he went on to explain Gauss's law this way: imagine a point charge with electric field around it $$E=\frac{q}{4\pi\varepsilon_0r^2}.$$ If we draw a spherical closed surface around the charge, the flux through the sphere is: $$ flux = \iint E \cos(\theta) dA,$$ where $E$ is on the surface, and $\theta$ is the angle between $E$ and the surface vector. Since $\cos(\theta)=1$ for all points, and $A=4\pi r^2$, then the flux is $q/\varepsilon_0$ (or how I understood it, the decrease of the field strength $E$ is offset perfectly by the increase of area $A$).

He then says that this law holds true for any closed surface. How is this possible? For a cubical surface, the $\cos(\theta)$ is not going to be $1$ everywhere, and the area is $24r^2 \implies \frac{q}{4\pi\varepsilon_0 r^2} \times 24r^2 = \frac{6q}{\pi \varepsilon_0}$? Why are cubes gaussian surfaces?

  • 1
    $\begingroup$ Tip: use laTex notation for formulas: everything you write between dollar symbols '\$ math text \$' will become a formula. Use laTex commands inside. $\endgroup$
    – FGSUZ
    Dec 30, 2017 at 22:52
  • $\begingroup$ +1 for a nice question and showing your thinking behind it, just one other "issue". People like paragraphs (as in the answer you received) and it really helps to make your question readable. BTW, welcome to PhysicsSE :) $\endgroup$
    – user179430
    Dec 30, 2017 at 23:08

2 Answers 2


Okay I think I understand your question. Yes, the area is different, but the angle is different too. When you consider both, you actually have a different area times a different angle, but if you compute that "area times angle cosine" you will get the projection of that area over the unit sphere, so it is equivalent.

See the formulas: you've got that a radial field in scalar product with the surface:

$ \frac{Kq}{r^2}\hat{r}\cdot d\vec{S}$

but this is

$ {Kq}\ \frac{\hat{r}\cdot d\vec{S}}{r^2} = Kq \ \ d\Omega$

So it is the solid angle. Integration gives the solid angle $4\pi$ because it is a closed surface as well. Since the constant is $\frac{1}{4\pi\varepsilon_0}$, you get that $4\pi$ times that quantity is $\frac{q}{\varepsilon_0}$, the same result. This is not surprising, because it doesn't depend on the srface shape.

In sum, the fact that the flux is a scalar product implies it is a projection. You can think about the projection on a sphere, so you've got the same problem, and so the same result and properties.


The best way to think about this is to ask: is the total flux through a cube which contains the sphere different than the total flux through the sphere? The answer of course is no: the total flux is the same (provided there is no source or sink in the space between the sphere and the cube).

In the case of the sphere, the flux is uniformly distributed on the surface. Not so in the cube: there are patches where there will be more flux than others, but Gauss’ law is about net (or total) flux, so this can depend only on the strength of the source.

Of course because of the geometry it’s more difficult to compute the net flux through a cube, which is why for sources with spherical symmetry we might as well compute the flux using a sphere, knowing the flux through the sphere is the same as the flux through the cube.

As an analogy imagine a light bulb. The total flux of light through any closed surface depends not on the shape of the surface but on the bulb: a 60W bulb will produce more light than a 20W bulb. It doesn’t matter if the lampshade is in shape of a (closed) sphere, (closed) cylinder or cube, the total flux of light through the shade will depend only on the bulb, not the shape of the lampshade.


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