Tesla Coil Surprises Inside a Faraday Cage

AI Thread Summary
Inside a Faraday cage, a person can safely touch the interior while exposed to high voltage from a Tesla coil because the electric charge distributes itself on the outer surface, preventing electric fields from penetrating the interior. This phenomenon occurs as electrons repel each other, moving to the edges of the conductor to maximize distance from one another. In electrostatics, the electric field inside a conductor is zero, as charges rearrange to counteract external fields. However, in a copper wire within a circuit, free electrons can flow through the interior, unlike in an isolated conductor. This discussion highlights the principles of electrostatics and charge distribution in conductors.
cragar
Messages
2,546
Reaction score
3
When this guy gets inside the Faraday cage and they hit it with like 200,000 volts from a Tesla coil , when he is inside the cage he can touch the inside and not get shocked , so basically the charge flows on the outside , is this because the electric field from all around the circle pushes the charge to the outside .?

 
Last edited by a moderator:
Physics news on Phys.org
I haven't heard many people talk about electric fields pushing!
A reasonable explanation coming from a slightly different direction is that electrons "don't like one another".
In their efforts to get as far away from one another as convenient, electrons on the inside of a conductive body get out onto the edges and even jump off into the atmosphere.

You might like the information on this link: http://www.howstuffworks.com/vdg.htm
 
thanks for the answer .
 
does the charge on a copper wire only flow on the outside of the wire ,
 
In electrostatics, the field inside a conductor is zero. By electrostatics, I mean an isolated conductor. If it is placed in an external field, the free electrons in the conductor would be acted upon by the field and would more towards the surface closer to the positive end, leaving behind positive atoms (or I should say ions). This separation of charges produces a field that opposes the external field. The separation takes place until the two fields cancel each other out. At this point the electron would be pulled by the the positive ions with the same force as the external force is pulling it in the opposite direction.

A copper wire, it is not an isolated conductor. It is in a circuit. The free electrons would be pulled out of the wire by the current source. As in the previous case they are not forced to settle onto the surface of an isolated conductor. So in a copper wire, electrons flow in the interior region as well.
 
ok thanks for the answer .
 
This is from Griffiths' Electrodynamics, 3rd edition, page 352. I am trying to calculate the divergence of the Maxwell stress tensor. The tensor is given as ##T_{ij} =\epsilon_0 (E_iE_j-\frac 1 2 \delta_{ij} E^2)+\frac 1 {\mu_0}(B_iB_j-\frac 1 2 \delta_{ij} B^2)##. To make things easier, I just want to focus on the part with the electrical field, i.e. I want to find the divergence of ##E_{ij}=E_iE_j-\frac 1 2 \delta_{ij}E^2##. In matrix form, this tensor should look like this...
Thread 'Applying the Gauss (1835) formula for force between 2 parallel DC currents'
Please can anyone either:- (1) point me to a derivation of the perpendicular force (Fy) between two very long parallel wires carrying steady currents utilising the formula of Gauss for the force F along the line r between 2 charges? Or alternatively (2) point out where I have gone wrong in my method? I am having problems with calculating the direction and magnitude of the force as expected from modern (Biot-Savart-Maxwell-Lorentz) formula. Here is my method and results so far:- This...

Similar threads

Back
Top