Having trouble understanding "Faraday cage"

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SUMMARY

The discussion centers on the principles of a Faraday cage, which is defined as a hollow metallic conductor that ensures an electric field inside is zero, represented by the equation $$\oint \vec{E}\cdot\,d\vec{A} = \dfrac{Q_{encl}}{\epsilon_{0}}$$. Participants clarify that a Faraday cage acts as an electrostatic conductor, where charges redistribute to cancel external electric fields. The conversation also distinguishes between conductors and electrostatic conductors, emphasizing that electrostatic conditions arise only after charge redistribution occurs within the conductor.

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  • Understanding of electrostatics and electric fields
  • Familiarity with metallic conductors and their properties
  • Knowledge of charge distribution in conductors
  • Basic grasp of the equation $$\oint \vec{E}\cdot\,d\vec{A} = \dfrac{Q_{encl}}{\epsilon_{0}}$$
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  • Research the properties of electrostatic conductors
  • Study the behavior of electric fields in conductors under various conditions
  • Explore the concept of charge redistribution in superconductors
  • Learn about the applications and limitations of Faraday cages in electromagnetic shielding
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Students of physics, electrical engineers, and anyone interested in the principles of electromagnetism and the practical applications of Faraday cages.

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Homework Statement


I am having trouble understanding how a Faraday cage works.

Homework Equations


$$\oint \vec{E}\cdot\,d\vec{A} = \dfrac{Q_{encl}}{\epsilon_{0}}$$

The Attempt at a Solution


It says that Faraday cage is a hollow metallic conductor and hence, inside the cage, $$\vec{E} = 0$$
I am aware that it holds for a eletrostatic metallic conductor, since if there is any electric field within the conductor, then the charges will move around by the field and it would be no longer electrostatic. Is Faraday cage a eletrostatic conductor as well? Is that why this still holds for Faraday cage? If it is, why is a Faraday cage electrostatic even if there is a electric field from outer source? Furthermore, I am wondering if all metallic conductors are naturally electrostatic and if they are, why. Thank you very much for your help.
 
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So why the charges on the surface of the cage move around to cancel the electric field from outside?
 
A Faraday cage is basically a conducting shell.
Suppose you have 3 of these shells (spherical for simplicity).
One has all of its excess charge on the outside.
Another has an equal distribution of its excess charges on both its inner and outer surfaces.
The third has all of its charge on its inner surface.
Which of the three has the lowest electrical potential?
Since charge is free to move about within a conductor it will assume the lowest energy configuration.
 
betelgeuse91 said:
So why the charges on the surface of the cage move around to cancel the electric field from outside?
You seem to be making some distinction between a conductor and an electrostatic conductor. (The metallic aspect is clearly irrelevant: it's a conductor.) Electrostasis is a circumstance, not a property of the material. If you drop a charge onto one spot on a conductor it will not immediately be electrostatic, it will take some fraction of a second for the charge to redistribute. (In fact, if it is a superconductor a current will continue to flow. It only settles down because of resistance.) Once redistributed, it will be electrostatic again.
If there is a field within the material of the conductor, it will push the free electrons around. They will only stop moving when there is no field.
 

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