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I was given the following explanation of Faraday cage: There is no electric field inside a hollow, charged conducting shell because a Gaussian surface inside the shell must have zero flux through it since it encloses zero charge. In the picture below, ${\bf S}$ is the Gaussian surface.

Charged shell

But I am not convinced by this explanation. There is clearly a field outside the shell since the shell is charged. What if I consider a closed surface ${\bf T}$ outside the Faraday cage like in the image below:

Another charged shell

Again, the charge enclosed by ${\bf T}$ is zero so shouldn't this mean that there is zero flux through ${\bf T}$ and hence no electric field inside ${\bf T}$?

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    $\begingroup$ Am I getting inconsistent results because I'm ignoring the direction of electric field? $\endgroup$
    – Confuse
    Nov 1, 2019 at 10:05

6 Answers 6

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All electric flux start at a positive charge (a.k.a source of flux) and end at a negative charge (a.k.a sink). Since the Gaussian surface you considered outside the charged shell houses neither positive nor negative charges, there can't be any net electric flux going into or coming out of the Gaussian surface of your choice.

That means all electric flux that goes in must come out.

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In the second case, the netto flux is also zero. There is an equal amount electric field going in the volume as out the volume. Hence, if you take the surface integral over the volume, the result is zero.

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  • $\begingroup$ But there is electric field inside T right? $\endgroup$
    – Confuse
    Nov 1, 2019 at 10:23
  • $\begingroup$ Yes, there will be electric field inside. $\endgroup$
    – Frederic
    Nov 1, 2019 at 10:25
  • $\begingroup$ Since there is no electric field inside S, I am assuming that the only reason why T has zero net flux despite of having a electric field inside is because of signs of the field lines (field lines entering and exiting have opposite signs) $\endgroup$
    – Confuse
    Nov 1, 2019 at 10:28
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    $\begingroup$ Yes, exactly. The electric field which enters the surface has opposite sign to the electric field which leaves the surface in the calculation of the surface integral. $\endgroup$
    – Frederic
    Nov 1, 2019 at 10:30
  • $\begingroup$ Thanks a lot for help! $\endgroup$
    – Confuse
    Nov 1, 2019 at 10:31
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Again, the charge enclosed by T is zero so shouldn't this mean that there is zero flux through T and hence no electric field inside T?

Let's say we have a point charge $q$ at the origin $O$. Let $S$ be any closed surface that doesn't contain $q$. Electric field is non-zero everywhere (It's $\vec{E}(\vec{r})= \frac{kq}{r^2} \hat{r}$ : we know this). And yet, what we find is that $\oint_S \vec{E} \cdot d\vec{S} = 0$ for such a closed surface even though electric field is non-zero inside $S$. The same applies to your scenario for the second closed surface you drew outside the cage.

So yes, as you say, the argument to show that the electric field inside a cavity of a conductor that contains no charge is zero, involves more reasoning than what has been told to you. Please refer to the following picture that is taken from Griffiths' Electrodynamics book.

enter image description here

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If T is also a hollow conductive shell then there is zero flux through T and hence no electric field inside T.

This happens because the external electric field, due to the charge distribution in S, produces a displacement of free charges in T (a conductor) which generates an charge re-distribution in T.

This new distribution of charge in T produces a local electric field in T which opposes to the external electric field caused by S. The result is zero net flux in T (equilibrium is re-established).

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If you have a conducting wire with all the charge is confined to flow within the (regardless of the potential across it).

Now, bend that wire into a loop. Is there current outside or inside of the loop?

What if you extend that into a 3d shell?

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There is a subject similar in hydraulic is about buoyancy. Where the vector to sum is pressure and the it said that it acts normal to the surface, but at the end it is a sum of forces.

Now for the electric field at a point it will give you the force for that possible test charge inside the metal of the surface. And it will be different at each point in the surface the electric value at the point near will be larger and the one who are far will be smallest. But the cumulative sum will be zero. enter image description here

I suppose that those forces will redistribute the charge in the Faraday cage but I don't know about that, I haven't read about that subject.

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