Gauss' law and Faraday cage problem

• sss1
sss1
Homework Statement
In the picture
Relevant Equations
Gauss's law
Is this a good response?
The lift is a conductor, therefore electrons can move freely. The charges on a conductor reside on the outer surface as they like to be as far from each other as they possibly can be due to the repulsive coulomb force. There is no charge between the inner and the outer surface of the box, the charge distribution on the outside of the box will not be uniform since its not spherical in shape. There will be more charges at the corners of the box. Using Gauss's law, drawing a rectangular Gaussian surface with width greater than the inner surface but smaller than the outer surface, the net enclosed charge inside the lift is 0 whether or not if there is charge on the student; the electrons inside the conductor will rearrange themselves to cancel out the electric field carried by the charge. Hence there is no electric field inside the lift, and so the radio does not work as it requires external electromagnetic signals. Outside the box, since negatively charged electrons reside on the surface, the electric field will terminate at the surface of the box.

I'm thinking of something like this for the diagram?

With regards to the charge distribution on the inner surface, doesn't that depend on whether or not the student has any charge? If the student carries some negative charge then there will be an induced positive charge around the inner surface; and there will be negative charges on the outer surface. Whereas if the student carries positive charge then there will be induced negative charge; and there will be positive charges on the outer surface.

I think the essence of this problem is that there’s no field inside the conductor (the enclosed metal box I.e. the elevator/lift).

Signal transmit through electric fields. If there is no field there is no signal.

My 2 cents.

To @PhDeezNutz's 2 cents, I will add my own to make a total of 4. Yes, the electric field is zero inside the conductor, but the magnetic field in the EM signal is also canceled by the Faraday's law eddy currents mostly on the outer surface. Read about the skin effect here.

PhDeezNutz
kuruman said:
To @PhDeezNutz's 2 cents, I will add my own to make a total of 4. Yes, the electric field is zero inside the conductor, but the magnetic field in the EM signal is also canceled by the Faraday's law eddy currents mostly on the outer surface. Read about the skin effect here.
so the signal cant propagate because the e field is being canceled out due to the electrons inside the conductor rearranging themselves and the magnetic field in the signal is canceled by Faraday's law eddy currents?

PhDeezNutz
sss1 said:
so the signal cant propagate because the e field is being canceled out due to the electrons inside the conductor rearranging themselves and the magnetic field in the signal is canceled by Faraday's law eddy currents?
Right. And if you have a dielectric like glass with no free electrons, light propagates through it with practically no problem. Have you ever wondered why a thin sheet of tin foil help up in front of a light bulb allows no light through whereas an equal thickness of paper allows quite a bit of light?

I'm a bit confused as to why the problem is asking the OP to explain this RF EM shielding effect using Gauss' Law though...

nasu
kuruman said:
Have you ever wondered why a thin sheet of tin foil help up in front of a light bulb allows no light through whereas an equal thickness of paper allows quite a bit of light?
well tin foil is conductor so electrons rearrange themselves and the e field from the light, since they're EM radiation, cannot propagate. So no light? Whereas paper is an insulator so EM radiation can move through?

Doesn't that make sense?

Generalizations of the type conductor is opaque and dielectric is transparent based on simplistic models are "dangerous". Glass may be opaque in some regions of infrared and ultraviolet and metals may be transparent in ultraviolet or x-rays.
Water absorbs microwaves even if it's pure (so not conductive).

Sure, but as a zeroth order explanation in the visible region with which most of us are familiar from daily experience is a good starting point.

kuruman said:
Doesn't that make sense?
What makes sense? The Tin foil and the light bulb?

What is Gauss' Law?

Gauss' Law states that the total electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the medium. Mathematically, it is expressed as ∮E·dA = Q_enclosed/ε₀, where E is the electric field, dA is a vector representing an infinitesimal area on the closed surface, and Q_enclosed is the total charge within the surface.

How does a Faraday cage work?

A Faraday cage works by redistributing electric charges on its surface in response to an external electric field, effectively canceling the field's effects inside the cage. This phenomenon is due to the conductive nature of the cage material, which allows free electrons to move and counteract the external electric field, thus shielding the interior from electric fields.

Can Gauss' Law be applied to a Faraday cage?

Yes, Gauss' Law can be applied to a Faraday cage. When considering a Gaussian surface inside the cage, the electric flux through the surface is zero because the net electric field inside the cage is zero. This implies that any external electric fields do not penetrate the cage, confirming its shielding effect.

Why is the electric field inside a Faraday cage zero?

The electric field inside a Faraday cage is zero because the conductive material of the cage redistributes its charges in such a way that it cancels any external electric fields. The induced charges on the surface of the cage create an opposing field that neutralizes the external field within the enclosed space, resulting in a zero net electric field inside.

A Faraday cage is effective at blocking static and low-frequency electric fields. However, its effectiveness at blocking higher-frequency electromagnetic radiation, such as microwaves or radio waves, depends on the size of the mesh or the material's conductivity. For very high frequencies, the cage may not provide complete shielding if the mesh size is comparable to the wavelength of the radiation.

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