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Suppose I have an infinitely tall solid cylinder with radius $R$ and charge density $\rho$.

enter image description here

Then, a portion of that cylinder is enclosed by a Gaussian surface, in this case another infinitely long cylinder with radius $r,\quad r<R$, as shown:

enter image description here

Suppose that $\rho$ only changes wrt $r$. Due to the symmetry, the electric field will therefore be constant along the Gaussian surface. Thus, $\Phi = EA$ where $A$ denotes the surface area.

Lastly, let $Q$ denote the charge enclosed in the Gaussian surface.

By Gauss' Law, I can state that the magnitude of the electric field along the Gaussian surface is $$E=\dfrac1A \dfrac {Q}{\epsilon_0}$$

Question. Why does the charge inside the Gaussian surface only contribute to the electric field? I would think that the charge outside the Gaussian surface also causes an electric field of its own, so why are the fields not superimposed?

EDIT: I am aware that I am dealing with a special case; that's why I ask the question. I can work out mathematically what the electric field would be. My question is, intuitively, why would charges outside the Gaussian surface offer no contributions to the electric field. It makes sense why, for instance, they offer no contribution to electric flux/charge (since they are outside the Gaussian surface, and thus the field lines cancel), but I still can't see why the electric field at the surface would be unaffected by outside charges.

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It's just a consequence of the symmetry of your problem and has nothing to do with Gauss' law per se. Gauss' law is just a simple way to extract the electric field in an easy way in the presence of a symmetry.

I will start with a brief explanation and then move on to a simpler example. At the very end, a small sketch clarifies why charges cancel.

Brief comment

Gauss' law tells you that only charges inside a surface contribute to the flux of the electric field on that surface. Intuitively, this is because the field due to charges inside crosses the surface only once while those due to charges outside crosses it twice, once entering and once leaving. The net effect is that the net flux of those charges is 0.

Now, in case of symmetry, like the one you mention, also the electric field due to charges outside cancels. Each charge outside produces a field on the inner cylinder but there are other charges that cancel it out. You could show it by summing the contribution of each charge outside separately but Gauss' law provides a faster answer.

This is not true for the charges inside which do not balance each other and have a net contribution.

However, this is an effect of symmetry, not an effect of Gauss' law. If you take an asymmetric situation, then, while charges outside do no contribute to the electric flux, they would contribute to the electric field.

Example

As a simple example, take two charges $q$ on the x axis, one in the origin and one in $x=a$, then take a spherical surface centered in the origin and with radius $R<a$.

The flux of the electric field on such a sphere would be $$\Phi(E)=q/\epsilon_0$$ i.e. only due to one charge inside. But now, because the situation is asymmetric, you can't find a simple expression for $\Phi(E)$ which allows you to related the field to the flux because you don't have symmetry anymore. In this case also the charge outside contributes to the electric field on the surface despite not contributing to the flux.

If now we try to recovery symmetry, by adding more charges, e.g. a spherical shell of charges of radius $a$ then now, because of symmetry again we can write $$\Phi(E)=EA$$ where $A=4\pi R^2$ is the surface of the sphere and you get $$E={1\over 4\pi R^2\epsilon_0}q$$ i.e. a field equal to the one simply due to the charge inside.

This means that now the charges outside balance their field and only the one inside has a net effect.

So we had the same situation: the same charge inside, the same surface and we only varied the distribution of charge outside. In both cases, $$\Phi(E)=q/\epsilon_0$$ i.e. only the charge inside contributes to the flux by Gauss' law.

However, only if the distribution outside is symmetric then to the charges outside balance each other's field. In the case of an asymmetric distribution, then the charges outside, while not influencing the flux, do influence the electric field on the surface.

Why do charges outside cancel

This needs to be worked out time after time, but the concept is that for any charge outside, in symmetric situation, you will find another one with the same but opposite electric field.

See this small drawing: the "orange" charge outside creates a big orange electric field on a point of the inner surface but that field is balanced by a lot of other yellow charges on the opposite side. This is valid for each point on the inner surface! Ths opposite would not be true, because the charges inside only go "radially" out so they can not balance each other and they end up all summing! But this is only because the system is symmetric! If you only had half the outside cylinder, the field would not cancel!

enter image description here

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  • $\begingroup$ Thank you for your answer. However, it seems to be contradicting the one by @YouFoundMe, who says the field I calculated was only the contribution of the charge enclosed within the Gaussian surface... $\endgroup$
    – user256872
    Mar 5, 2021 at 9:56
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    $\begingroup$ I disagree with that interpretation. In your example, your $E$ IS the net electric field in that point. It is the sum of all electric fields of all charges and it turns out only those inside contribute with a net effect (see drawing). However, in general, the flux can not simply and always be written as $\Phi(E)=EA$ which is only valid the the net field is perpendicular to the surface of your choosing. And indeed in those cases Gauss' law is not useful for finding the net field. But in your symmetric case, you are actually find the net total $E$! $\endgroup$
    – JalfredP
    Mar 5, 2021 at 10:00
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In general, gauss's law looks like $$\int_\mathcal{S}\mathbf{E}\cdot \mathbf{d}S=\frac{q_\text{enc}}{\epsilon_0}$$ Which says in words, that the flux of electric field through a closed surface is equal to $1/\epsilon_0$ times charge enclosed by the volume. If you go through the detail, it will make sense.


This law doesn't say that the charge that is outside the surface can't contribute to the field but it's said that they can't contribute to the flux of the field through the surface. In general $$\int_\mathcal{S}\mathbf{E}\cdot d\mathbf{S}\not=\mathbf{E}\int_\mathcal{S} d\mathbf{S}$$ But under certain symmetric conditions like the one you came up with. We can do this. If that is the case then $$\mathbf{E}=\frac{q_\text{enc}}{\epsilon_0}\frac{1}{\int_\mathcal{S} d\mathbf{S}}\ \ \ \text{under certain symmetric conditions}$$ In these cases, the outside charges don't contribute to the field.

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  • $\begingroup$ I am aware that I am dealing with a special case; that's why I ask the question. I can work out mathematically what the electric field would be. My question is, intuitively, why would charges outside the Gaussian surface offer no contributions to the electric field. It makes sense why, for instance, they offer no contribution to electric flux/charge (since they are outside the Gaussian surface, and thus cancel), but I still can't see why the electric field would be unaffected by outside charges. $\endgroup$
    – user256872
    Mar 5, 2021 at 9:20

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