Electronics > Metrology

Coulomb's law and a voltage frame of reference

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John Heath:
In special relativity the laws of physics are the same for all frames of reference. Can one assume from this that laws of physics are the same for all voltage frames of reference? I will charge a Faraday cage to 10 K volts positive then enter it. From within the 10 K volt cage there is no experiment or measurement that can be made to indicate the voltage frame of reference is 10 K volts above earth. Can not reach outside the Faraday cage with a voltage probe as that would cheating. Must be from within the Faraday cage. Is there any way to tell?

Hair should start to stand out at 10 K volts as like charges repel. However the entire Faraday cage is 10 K volts so it cancels out. Yes and no. Yes it cancels out but Coulomb's law of like charges repelling is Q x Q / R^2 = force where Q is charge and R is distance. R^2 is the problem. The hairs on my head are 1/10 of an inch apart but the Faraday wall is 5 feet away. According to Coulomb's law my hair should be standing out. However if my hair were standing out I would be in violation of the said postulate that the laws of physics are the same for all voltage frames of reference as I would know that my Faraday cage is 10 K volts above earth ground. On the other hand if my hair does not stand out then we are in violation of Coulomb's law Q x Q / R^2 = force. Therein is the rub.

I can venture a guess that my hair would not stand out in this Faraday experiment as an airplane at 32 thousand feet has a sky charge of 50 to 100 K volts. If hair were standing out on airplanes it would be known. This leaves Coulomb's law of Q X Q / R^2 = force in question with focus on R^2 distance . Then again Coulomb's law is known to be true.

If it helps charge a classic gold leaf to 10 K volts then seal it electrically in a glass jar. It should be standing out from a Coulomb force of 10 K volts. Then place the sealed gold leaf inside the 10 K volt charge Faraday cage. It should still be standing out with Q X Q / R^2 = force but if it did the one inside the Faraday cage would be in violation of the laws of physics being then same in all voltage frames of reference as he would know he is in a different voltage frame of reference? Any thoughts on this would be appreciated.

ruffy91:
If you bring the jar inside the cage you change the frame of reference. It's still charged 10kV above reference (So cage is at 10kV and gold at 20kV -> still same force as potential difference is still 10kV)

Andy Watson:

--- Quote from: John Heath on April 24, 2016, 12:49:50 pm ---Must be from within the Faraday cage. Is there any way to tell?

--- End quote ---
No way to tell for a complete Faraday cage. Experimentally, the absence of a detectable E field within an enclosed conducting surface is used to prove the inverse square law. Rather than say that like charges repel, if you consider the action of a single charge to be the result of interacting with the E field, once you make the E field zero there is no force on the charge.

Bleaney and Bleaney give description of your Faraday cage problem under "Experimental proof of the inverse square law" - chapter 1.
https://archive.org/details/ElectricityMagnetism2ndEd

Marco:
Mass of the cage would change slightly.

orolo:
A Faraday cage is charged. Inside, owing to Gauss' Law, there is no E field. You are inside and start rotating around yourself in place. That is equivalent to the charges in the cage rotating around you in the opposite direction: moving charges cause a magnetic field. So you should be able to measure a magnetic field while spinning. In other words, a rotating charged sphere should have a measurable magnetic moment.