Author Topic: Coulomb's law and a voltage frame of reference  (Read 21137 times)

0 Members and 1 Guest are viewing this topic.

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Coulomb's law and a voltage frame of reference
« on: April 24, 2016, 12:49:50 pm »
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.

 

Offline ruffy91

  • Regular Contributor
  • *
  • Posts: 240
  • Country: ch
Re: Coulomb's law and a voltage frame of reference
« Reply #1 on: April 24, 2016, 01:25:44 pm »
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)
 

Online Andy Watson

  • Super Contributor
  • ***
  • Posts: 1853
Re: Coulomb's law and a voltage frame of reference
« Reply #2 on: April 24, 2016, 01:42:43 pm »
Must be from within the Faraday cage. Is there any way to tell?
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
« Last Edit: April 24, 2016, 01:52:19 pm by Andy Watson »
 

Offline Marco

  • Super Contributor
  • ***
  • Posts: 5637
  • Country: nl
Re: Coulomb's law and a voltage frame of reference
« Reply #3 on: April 24, 2016, 01:47:58 pm »
Mass of the cage would change slightly.
« Last Edit: April 24, 2016, 01:49:40 pm by Marco »
 

Offline orolo

  • Frequent Contributor
  • **
  • Posts: 352
  • Country: es
Re: Coulomb's law and a voltage frame of reference
« Reply #4 on: April 24, 2016, 02:00:24 pm »
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.
 

Online T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 19310
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: Coulomb's law and a voltage frame of reference
« Reply #5 on: April 24, 2016, 02:49:42 pm »
Mass of the cage would change slightly.

Hmmm, due to the difference in electrons?  Possibly.

There will also be a change in mass due to the energy stored in the electric field, but this is a property of the space itself where those fields exist, not just of the cage.  Indeed, there's no field and thus no mass change from within the cage.

I suppose one could determine the mass change "from the inside" by shaking the cage -- assuming it's on a sufficiently compliant suspension, or say, floating in space.  But this wouldn't uniquely identify the mass change; it could be due to other fields or effects instead.

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #6 on: April 24, 2016, 03:41:43 pm »
Must be from within the Faraday cage. Is there any way to tell?
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

Let us set some limits. The stray capacity when charging the gold leaf was 10 p Farads and the current to charge the gold leaf to 10 K volts was 1 p amp. 1 amp is  6.241×10^18 electrons so 1 p amp is 6.24 X 10^6 electrons. The gold leaf is missing 6.24 X 10^6 electrons  and electrically sealed to prevent electrons from returning. Let us also also set the stray capacity of the gold leaf in the Faraday cage at 10 p Farad for simplicity. With all this in place I would have to agree with ruffy91. All we did was change the voltage reference from 0 to 10 K volts therefore the gold leaf is now 20 K volts. Your detectable E field , voltage gradient , is between the Faraday wall and the gold leaf that is electrically sealed and still missing 6.24 X 10^6 electrons when I charged it outside the Faraday cage. Did not mean to complicate the issue by counting missing electrons , 6.24 X 10^6 , but Coulomb's laws is all about charges not voltage where Q in Q X Q / R^2 is number of charges , number of missing electrons , not the voltage gradient. Can I say that +10 K volts and + 10 K volts relative to ones voltage frame of reference always equals a given repulsive force at distance X or do we have to stick to Coulomb's law that charge parity , accounting of number of extra electrons or missing electrons equals force with Q X Q / r^2= force ?? I am not sure. If I had a infinitely high impedance meter to measure the  voltage of 1 electron then yes maybe. What is the voltage of one electron relative to an infinite number of match electrons to protons such as earth?

There is a larger issue here that I did not want to bring up as it just complicates it too much. I am going to bring it up anyways, apologies in advance. An electron and positron have the exact same mass. If I drag both Mr positron and electron into a 10 K volt + charged Faraday cage the electron will go in willingly but the positron will resist as positive charge do not like a 10 K volt Faraday cage. This means the positron must be forced into the positive Faraday cage but the electron not. For energy conservation laws we have to reduce the mass of the electron and increase the mass of the positron to justify this. Now energy is conserved. The chances of earth being a perfect 0 voltage frame of reference are extremely unlikely so how could we have a electron and positron of precisely the same mass ?? No answers on my end. I am just saying there is something fishy about an electron and positron having the exact same mass.     
 

Online T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 19310
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: Coulomb's law and a voltage frame of reference
« Reply #7 on: April 24, 2016, 04:05:08 pm »
The chances of earth being a perfect 0 voltage frame of reference are extremely unlikely so how could we have a electron and positron of precisely the same mass ?? No answers on my end. I am just saying there is something fishy about an electron and positron having the exact same mass.   

The Earth is pretty damn near neutral, with, I'd guess, at most a few billion volts potential with respect to the interplanetary medium.  Which isn't much electric field on a planetary scale (there's as much voltage in the atmosphere itself, occasionally building up to breakdown, causing teeny little sparks.. :) ).  Probably a physicist has ran the numbers and come up with a more reasonable (and even smaller) quantity; for example, photoemission isn't going to push more than x eV of voltage; the highest energies coming from the Sun are in the 10MeV range (coronal electrons), which would simply get trapped if there were much more field nearby.

In any case, energy is stored in the field, not the particle (except as a matter of bookkeeping), so once you've performed the work to put the particles inside a given region, the particles themselves don't "know" anything further.

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline orolo

  • Frequent Contributor
  • **
  • Posts: 352
  • Country: es
Re: Coulomb's law and a voltage frame of reference
« Reply #8 on: April 24, 2016, 04:20:05 pm »
You can easily introduce a positron inside an arbitrarily charged positive farady cage: just introduce a ~1MeV photon inside, and let it decompose into an electron-positron pair, and then make the electron hit the cage (just a gentle push, there is no E field), changing the cage's charge by a trivial amount, while you retain the positron. Now the positron is inside, no problems at all. This might seem contrived, but makes perfect physical sense. Besides, any excess of mass attributed to positrons should be also attributed to electrons, reversing the experiment with a negatively charged cage. Of course the electron and the positron have the same mass, they are the same particle after all: they are both eigenstates of the same Dirac equation.

Quote
What is the voltage of one electron relative to an infinite number of match electrons to protons such as earth?

Assuming a perfectly conducting Earth, it's just the complementarity principle: the field is equivalent to that generated by an opposite charge simmetrically placed with respect to earth.
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #9 on: April 24, 2016, 04:59:13 pm »
Mass of the cage would change slightly.

Hmmm, due to the difference in electrons?  Possibly.

There will also be a change in mass due to the energy stored in the electric field, but this is a property of the space itself where those fields exist, not just of the cage.  Indeed, there's no field and thus no mass change from within the cage.

I suppose one could determine the mass change "from the inside" by shaking the cage -- assuming it's on a sufficiently compliant suspension, or say, floating in space.  But this wouldn't uniquely identify the mass change; it could be due to other fields or effects instead.

Tim

Maybe you are on to something . Mass change means frequency change. Frequency change means a change to either inductance or capacity or both of a L/C resonate circuit. If Uncle Albert has it right then it is the inductance L that changed for the relative inertia of a system to cause time to dilate , lower L/C resonance frequency . On the other hand if capacity changed then it is a different story. A fork in the road. Which way to turn. There is a possible solution found in the work of Oliver Heaviside to calculate impedance. If impedance is higher then a reflected pulse will be the same polarity. If the impedance is lower then the reflection is reversed in polarity.  The math works out that you know if it was L or C that changed if the impedance change is known provided the speed of light is constant. Speed of light becomes 1/resonate frequency for vacuum properties 8.8 p f and 1.28 u Henry and impedance Z 376 ohms. If relativity is right then the impedance should be Z 377 for an increase in relative inertia of a system  or the alternative 375 Z ohms for an increase in the relative size of the system , capacity C. If the 10 K volt Faraday cage door could be left open a jar then a test could be done to see if a EM pulse reflects polarity the same or reversed to test for relative inertia of a system L vs relative size of a system C for a L/C resonate time dilation.
 

Offline Zeranin

  • Regular Contributor
  • *
  • Posts: 179
  • Country: au
Re: Coulomb's law and a voltage frame of reference
« Reply #10 on: April 24, 2016, 08:25:02 pm »
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.

I have thought about the same thing long ago, and concluded that that there IS such a thing as a an 'absolute voltage', though an electrometer won't measure it if inside a Faraday cage at the same potential. In such a case, there will be no electric fields anywhere within the cage, because the cage and all within are at the same potential, so the electrometer won't register, but Coulomb's law is still not violated. The gold leaves are still being repelled by each other, but see an equal and opposite force on account of the charged cage. I'll think further about that, but a first glance that appears to be the case.This discussion leads naturally to asking, then what is the absolute voltage of the Earth, and what keeps it in check, and from memory the answer to that lies in the 'solar wind'. I do a lot of work calculating the electric fields in complex geometries in particle accelerators and the like, so have some familiarity with this kind of stuff. I've gotta go for now, but will think further about your interesting questions.
« Last Edit: April 24, 2016, 09:13:31 pm by Zeranin »
 
The following users thanked this post: John Heath

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #11 on: April 25, 2016, 01:47:50 am »
You can easily introduce a positron inside an arbitrarily charged positive farady cage: just introduce a ~1MeV photon inside, and let it decompose into an electron-positron pair, and then make the electron hit the cage (just a gentle push, there is no E field), changing the cage's charge by a trivial amount, while you retain the positron. Now the positron is inside, no problems at all. This might seem contrived, but makes perfect physical sense. Besides, any excess of mass attributed to positrons should be also attributed to electrons, reversing the experiment with a negatively charged cage. Of course the electron and the positron have the same mass, they are the same particle after all: they are both eigenstates of the same Dirac equation.

Quote
What is the voltage of one electron relative to an infinite number of match electrons to protons such as earth?

Assuming a perfectly conducting Earth, it's just the complementarity principle: the field is equivalent to that generated by an opposite charge simmetrically placed with respect to earth.

Your thoughts brought up an interesting idea. If I pair separate +e and -e out of the vacuum with your 1 MeV spark plug the positron will not last long before it annihilates with the plentiful electrons in the Faraday cage. I will make the Faraday cage more and more negative until the life expectancy of the positron equals the life expectancy of the electrons. That could be more and more positive depending on how you think of it. In any event when pair separated -e and +e reaches equilibrium 50/50 for life expectancy the voltage of the Faraday cage should be the zero voltage frame of reference of the universe ,, maybe. In short the laws of physics are not the same for all voltage frames of reference if Mr positron has a very short life compared to the electron. The positron being anti matter in equilibrium with electron , matter , makes this an experiment best done at a distance with sun block 8)
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #12 on: April 25, 2016, 04:05:09 am »
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)

If the leaf touches the inside Faraday cage it will not discharge. If it will not discharge then there can not be a voltage difference between the inside walls and the charged leaf. The person inside the cage can not witness a separated gold leaf if  he measures 0 voltage difference between the leaf and the Faraday cage walls as his laws of physics must be the same as ours. It gets worse. If he walks outside the Faraday cage he may discharge the gold leaf to the outside of the Faraday cage. As strange and counter intuitive as that sounds it appears to be true. I have a video of such an experiment by Professor Eric Rogers. He has a propensity to electrocute little girls but his physics is cool.



 

Online T3sl4co1l

  • Super Contributor
  • ***
  • Posts: 19310
  • Country: us
  • Expert, Analog Electronics, PCB Layout, EMC
    • Seven Transistor Labs
Re: Coulomb's law and a voltage frame of reference
« Reply #13 on: April 25, 2016, 04:22:25 am »
Whoa whoa, slow down there, John!

I'm not sure what sense to make of your other post, so about this one,

Your thoughts brought up an interesting idea. If I pair separate +e and -e out of the vacuum with your 1 MeV spark plug the positron will not last long before it annihilates with the plentiful electrons in the Faraday cage. I will make the Faraday cage more and more negative until the life expectancy of the positron equals the life expectancy of the electrons. That could be more and more positive depending on how you think of it.

All the positron sees, inside the cage, is the electric field near it.  If there is a net force, it accelerates in that direction.

If the field inside a Faraday cage weren't independent of outside fields (which is the definition of such a cage!), for example the drift region in any CRT would not work.

The last grid/anode in a CRT electron gun (in the neck) is the highest voltage anode.  After that point, the electric field is nearly zero -- the entire volume of the picture tube.  Most CRTs would not work if this weren't the case!

(Tektronix scope CRTs, for example, cheat by accelerating the beam with a relatively low potential (~2kV), deflecting it, then passing through a pair of screens which accelerate at a much higher potential (10-15kV).  Some of the beam strikes the screens, reducing beam current and scattering it, making the spot size larger, so it comes at a cost.  Still, the final drift region, between second screen and phosphors, is essentially a Faraday cage.)

So in conclusion, the net force on a positron (or electron) in the middle of a Faraday cage, is zero.  If the particle has nonzero velocity, it will eventually hit a wall, so this isn't a very effective trap.

If you can perform this measurement, with extremely large fields outside the cage, and do measure a difference (that cannot be attributed to leakage of fields through the cage), you will have some interesting physics papers to write!

Tim
Seven Transistor Labs, LLC
Electronic design, from concept to prototype.
Bringing a project to life?  Send me a message!
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #14 on: April 25, 2016, 11:36:54 am »
I see your point., Once the electron passes the neck of the CRT its in a Faraday cage and the voltage is not changing so the electron is coasting from that point on.
 

Offline orolo

  • Frequent Contributor
  • **
  • Posts: 352
  • Country: es
Re: Coulomb's law and a voltage frame of reference
« Reply #15 on: April 25, 2016, 02:38:51 pm »
If the leaf touches the inside Faraday cage it will not discharge. If it will not discharge then there can not be a voltage difference between the inside walls and the charged leaf.
Hi. If the gold leaf is charged, it will discharge when contacting the cage, regardless of the charge already stored in the cage: due again to Gauss' law, all the charge in the leaf will migrate to the outside of the cage. So even if the leaf discharges, it will tell nothing about the total charge of the cage.

This is beautifully demonstrated in a Van de Graaff generator. The VdG generator also exemplifies pumping positive charges inside an already positively charged (quasi perfect) faraday cage.
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #16 on: April 25, 2016, 10:13:59 pm »
Just curious -- why is this thread in the "Metrology" forum?   :-//

The jest of this thread is measurement. How to measure one's voltage frame of reference compared to the universe. Metrology seems like the  place to bring it up.
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #17 on: April 25, 2016, 10:46:35 pm »
If the leaf touches the inside Faraday cage it will not discharge. If it will not discharge then there can not be a voltage difference between the inside walls and the charged leaf.
Hi. If the gold leaf is charged, it will discharge when contacting the cage, regardless of the charge already stored in the cage: due again to Gauss' law, all the charge in the leaf will migrate to the outside of the cage. So even if the leaf discharges, it will tell nothing about the total charge of the cage.

This is beautifully demonstrated in a Van de Graaff generator. The VdG generator also exemplifies pumping positive charges inside an already positively charged (quasi perfect) faraday cage.



If you watch the video by Professor Eric Rogers it can be seen that the charged gold leaf will not discharge inside the Faraday Cage . outside yes but not inside. It is towards the end of the video.

Van De Graaf generator. I have a funny story about this that turned out to be a learning experience.  I bought a kids Van De Graaf just to try it out. It was cheap and I could see that copper wires at the bottom were not even toughing the belt. Good thing it was in my hands as I sorted that problem out making sure the belt had good contact with the copper brushes. This was not enough as it still did not seem to work properly. In a moment of desperation I broke the cardinal rule. I am not proud of this but I read the manual. I know I know but I really wanted to see it work. It said in rather large print " THE VAN DE GRAAF COPPER WIRES MUST NOT TOUCH THE BELT DIRECTLY" . I made the adjustments and it started to work , ha. Turns out Mr Van De Graff is in the business of moving protons not electrons and the path of least resistance for a proton is ionized air not copper. As the saying goes you are never too old to learn something new.
 

Offline Zeranin

  • Regular Contributor
  • *
  • Posts: 179
  • Country: au
Re: Coulomb's law and a voltage frame of reference
« Reply #18 on: April 26, 2016, 03:43:43 am »
I have thought further about this problem. Previously I said that an electrometer within the charged Faraday cage will not register that the cage is charged, and that is correct, but the explanation I originally gave is not. I'm still thinking about exactly what is going on, and the best way to explain it. I'll be back.  :)
« Last Edit: April 26, 2016, 06:39:40 am by Zeranin »
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #19 on: April 26, 2016, 12:02:48 pm »
Let me add to your thoughts with some out of the box ideas. Electron mass is known to be .511 MeV and proton 938.272 MeV. In a Faraday cage charged to - 1 MeV is this still true? To conserve energy I will say electron  .5 MeV and proton 939 MeV mass. Let us go more and more negative in the Faraday cage until the electron mass equals the proton mass at about 450 MeV each. This could be the point of absolute zero voltage frame of reference making the electron and proton the compliment of the same particle but opposite charge. That was a Paul Dirac idea. He later tossed it in the trash bin when the positron was discovered. I sometimes wonder if he was too quick to give up on this?   
 

Offline Zeranin

  • Regular Contributor
  • *
  • Posts: 179
  • Country: au
Re: Coulomb's law and a voltage frame of reference
« Reply #20 on: April 26, 2016, 12:15:42 pm »
Before embarking on an explanation, I need to explain what has been bothering me personally, about this problem. In a hand-waving way, a gold leaf electroscope measures how ‘charged’ an object is. If an object has an excess of charge (AKA the object is charged) then when connected to an electroscope, the gold leaf will rise through electrostatic repulsion of like charges. We all know that. OK, so we charge up a hollow sphere or Faraday cage, to say 10kV above ground, and connect our electroscope to the outside of the sphere, and of course the gold leaf flips up, confirming the fact that the sphere is charged. The entire sphere is at the same potential, because it is a metallic conductor. We then place an uncharged electroscope inside the charged sphere, and connect the electroscope terminal to the inside surface of the sphere, and it registers nothing, yet the inside and outside surfaces of the sphere are both at the same potential. How so? It does not need to be an electroscope, human hair standing on end performs the same experiment. Apparently we cannot detect or measure the charge on the sphere, if the observer or experiment is placed within the sphere, at least with an electroscope.

I believe that much of the confusion (or at least my confusion) stems from a misunderstanding of what and how an electroscope actually measures. As we know, an electroscope is a single terminal device. If the gold leaf is uncharged (equal number of +ve and –ve charges), then there is no repulsive force on the leaf, and no deflection of the leaf. If we push charge into the electroscope so that the leaf is ‘charged’, then the leaf will deflect.

That said, an electroscope can also function as a crude and insensitive voltmeter, which I would argue is how it is actually being used in these discussions. How so, when an electroscope has only one input terminal, while a voltmeter is a differential device with two input terminals, that measures potential difference? As I understand the original posting, the Faraday cage is charged up to 10 kV, meaning that a 10kV power supply is connected between Earth, and the cage. A wire is then taken from the cage to an electroscope, which will deflect. The deflection is proportional to the 10 kV, and the capacitance from the electroscope to Earth. The charge that will flow into the electrometer (which is what makes it deflect) is given by Q=CV, so for a given 10kV, the electrometer deflection will be proportional to this capacitance. That is why when an electroscope is used in this way, the case is grounded, to increase this capacitance. The net effect of all this is that the electroscope is actually measuring a differential voltage between it’s input terminal and case, just as with a conventional differential voltmeter.

Now we consider the case where the electroscope is placed inside the Faraday cage, which for convenience can be a spherical conducting shell. We connect the electroscope input terminal to the inside surface of the sphere/cage, but to what do we connect the electrometer case? The only thing we can connect it to is the sphere/cage, but if we do that then there will be zero volts between the input and case as they will be connected to the same thing, and we will get zero reading. We could leave the case unconnected, but that would be even worse, decreasing the capacitance from the electroscope to the cage. It’s best to forget about using an electroscope voltmeter at all, and just bring your favourite voltmeter into the cage, but you will have no more success in detecting that the cage is charged and by how much, for the only way to do that is to measure voltage between the sphere and ground, which you fundamentally cannot do if you and your equipment are inside the cage. Getting back to the electroscope, the only way that you can push charge into the electroscope and make it deflect, is if you have a potential difference, but there are no potential differences or electric fields to be found inside the cage, so the electroscope is never going to deflect.           

So what about the suggestion that we could detect the mass of excess charge on a metallic object within the sphere, that had touched the inside surface. That won’t work either, because all of the excess charge is distributed on the outside surface of the sphere/cage, so can’t be ‘seen’ or detected from inside the cage. To put that another way, as your mate outside the cage starts cranking up the 10kV power supply, at all times there are no electric fields anywhere inside the cage, so no driving force to push charge around so as to create an excess charge on anything inside the cage. Or deflect an electroscope or voltmeter inside the cage. I believe the OP is right. From within the cage, there is no way of knowing whether the cage is charged or not, or by how much.
 
From outside the cage it’s easy to detect if the cage is charged. If a large, electrically neutral body like Earth is conveniently nearby then use a voltmeter. If the cage is in deep space, then observe the motion of nearby charged particles.

Interesting topic, even if the link to metrology is tenuous.
« Last Edit: April 26, 2016, 12:34:42 pm by Zeranin »
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #21 on: April 26, 2016, 11:28:20 pm »
Some fine thoughts. Enjoyed the read.  Yes the electroscope has only one lead plus the stray capacity from the outside of the gold leaf to whatever is round. Let us make some differential voltage measurements then apply Coulomb's law outside the Faraday cage. From a grounded water tap the gold leaf measures 10 K volts. From the same water tap the chair , ceiling , floor and walls measure 0 volts. Let us apply Coulomb's law. Q x Q / R^2 = force. 0 X 10 k volt = 0 so the attractive force between the room and the gold leaf is 0. What about the leaf/ It measures 10 K volt on either side so 10K X 10 K / R^2 is an impressive repelling force.

What if we lived on the leaf and our water tap ground reference was the leaf itself. Without bias we will let the numbers tell us armed only with Coulomb's law. Relative to the leaf the walls and ceiling are - 10 K volts but they are far away and in every direction. Q x Q / R^2 so there is a small attractive force outward in all directions. What about the Leaf relative to the leaf?. The leaf relative to the leaf the voltage is 0 volts therefore no force to separate the leaf.

In summary we in the kitchen with our Coulomb's law book predict a strong outward force on the gold leaf. Our friends that live on the gold leaf armed with the same Coulomb's law book predict a weak outward force on the leaf.

There can not be two different prediction from the same theory of Coulomb's law yet there it is depending on the observer being in a kitchen or an observer living on a 10 K volt gold leaf. This is the part that keep me pacing the floor at night.
 

Offline Zeranin

  • Regular Contributor
  • *
  • Posts: 179
  • Country: au
Re: Coulomb's law and a voltage frame of reference
« Reply #22 on: April 27, 2016, 02:42:09 am »
Some fine thoughts. Enjoyed the read.  Yes the electroscope has only one lead plus the stray capacity from the outside of the gold leaf to whatever is round. Let us make some differential voltage measurements then apply Coulomb's law outside the Faraday cage. From a grounded water tap the gold leaf measures 10 K volts. From the same water tap the chair , ceiling , floor and walls measure 0 volts. Let us apply Coulomb's law. Q x Q / R^2 = force. 0 X 10 k volt = 0 so the attractive force between the room and the gold leaf is 0. What about the leaf/ It measures 10 K volt on either side so 10K X 10 K / R^2 is an impressive repelling force.

What if we lived on the leaf and our water tap ground reference was the leaf itself. Without bias we will let the numbers tell us armed only with Coulomb's law. Relative to the leaf the walls and ceiling are - 10 K volts but they are far away and in every direction. Q x Q / R^2 so there is a small attractive force outward in all directions. What about the Leaf relative to the leaf?. The leaf relative to the leaf the voltage is 0 volts therefore no force to separate the leaf.

In summary we in the kitchen with our Coulomb's law book predict a strong outward force on the gold leaf. Our friends that live on the gold leaf armed with the same Coulomb's law book predict a weak outward force on the leaf.

There can not be two different prediction from the same theory of Coulomb's law yet there it is depending on the observer being in a kitchen or an observer living on a 10 K volt gold leaf. This is the part that keep me pacing the floor at night.

 Let me summarize your situation, just to be sure I understand. You have a gold-leaf electroscope on your kitchen table, and you have a 10kV power supply connected between your water tap and the electrosope. You analyse the separating force on the gold leaf according to Coulomb's Law, from the perspective of sitting on your kitchen chair, and from the perspective of living on the gold leaf, and you get two different answers. Yes, that would  keep me awake at night also.

You seem to be mixing up ‘charge’ and ‘voltage’, in Coulomb’s Law, and be not paying attention to what the electroscope actually measures. You also seem to be forgetting that while voltage is nearly always described as a relative quantity, AKA potential difference, excess charge (the Q in Coulombs Law) is absolute, not relative. You can say that Earth is at some particular large voltage with respect to something else in the Universe, but this in no way alters the fact that the Earth is electrically neutral, Q=0. I’m also uncomfortable with your term ‘the leaf relative to the leaf’ . Both leaves are always at the same potential, because they are metal and connected to each other, but I’m sure you know that, so on that point I’m probably just arguing with semantics.
 
Fundamentally, the electroscope responds to charge, more specifically, excess charge on the leaf. It can be made to respond to a voltage, but only if and because such a voltage has the effect of producing excess charge on the leaf.

Here is how I analyse your example, in a way that produces consistent results regardless of the position and perspective of the observer. Assume the electroscope is of the type that uses a pair of gold leaves, that hang down vertically. There will be a stray capacity, ‘C’,  from each gold leaf to earth. You could increase C significantly by connecting your metal electroscope case to the water pipe, but either way, you have a particular stray capacity C from each leaf to ground. The fixed-voltage power supply is also connected from ground to leaves, so the power supply charges up this capacitor. Assuming that mother Earth is electrically neutral, the total excess charge that is pushed onto each gold leaf is given by Q=CV. The gold leaves then respond according to F = Q x Q / R^2, where Q is the charge on each leaf, and R is the separation between the leaves. This analysis and result is exactly the same, regardless of the observer, isn’t it? The stray capacity depends only on the geometry, which is observer independent, as is the magnitude of the power supply voltage, 10kV.

I’m not sure if the above explains what was bothering you, but if it does not, get back to me and we’ll thrash it out.
« Last Edit: April 27, 2016, 03:00:28 am by Zeranin »
 

Offline John Heath

  • Frequent Contributor
  • **
  • Posts: 444
  • Country: ca
  • 2B or not 2B
Re: Coulomb's law and a voltage frame of reference
« Reply #23 on: April 27, 2016, 04:22:45 am »
I like you model better. Let us make it clearer by floating in the middle of space somewhere. Stray capacity is no longer an issue. I will replace gravity with a spring on the gold leaf. If charged the leaf spring opens. Take the charge away and the leaf closes. Both us as observers from a distance and the ones that live on this gold leaf apply Coulomb's law as an accounting issue of literally counting the number of + charges and _ charges , protons / electrons , and applying Coulomb's law accordingly. This resolves the conflict between the distant observer and the one that lives on the gold leaf as they will now arrive at the same answer for the gold leaf.

I do not mean to be a trouble maker here but there is one fly left in the soup. Replace the spring gold leaf with a Faraday cage that has 0 charge. We place one electron in the middle somewhere. As the cage has 0 charge then Q X Q / R^2 or 0 X 1 electron = 0 force. The electron is free to float around not effected by a Coulomb force.  I am going to throw a wrench in the works by sprinkling some electrons on the Faraday cage. let us say the cage now has a charge of -1 and the electron floating inside has a charge of -1 just to give it a number. If Q X Q / R^2 is true then the electron should stay in and exact center of the cage as R^2 means closer to any side of the cage will increase the Coulomb force repelling it. The problem is the electron could care less and will still just float around indifferent to R^2 Coulomb force from the sides of the Faraday cage. This was demonstrated in the video by Professor Eric Rogers ironically using the inverse law R^2 to explain it. Maybe I misunderstood it.

He hints that it is similar to gravity at the center of the world where one weighs 0 pounds. That can be understood. However if the earth were a egg shell and we were inside the egg shell gravity would not pull one towards the shell just like the 1 electron floating around in a positive charged Faraday cage for this example would not feel a Coulomb force to the cage walls . How can an electron not want to go to a positive charged Faraday wall? In an oscilloscope the electron will go toward a positive deflection plate but not in the Faraday cage despite R^2. 

 
 

Offline Zeranin

  • Regular Contributor
  • *
  • Posts: 179
  • Country: au
Re: Coulomb's law and a voltage frame of reference
« Reply #24 on: April 27, 2016, 06:28:56 am »
I like you model better. Let us make it clearer by floating in the middle of space somewhere. Stray capacity is no longer an issue. I will replace gravity with a spring on the gold leaf. If charged the leaf spring opens. Take the charge away and the leaf closes. Both us as observers from a distance and the ones that live on this gold leaf apply Coulomb's law as an accounting issue of literally counting the number of + charges and _ charges , protons / electrons , and applying Coulomb's law accordingly. This resolves the conflict between the distant observer and the one that lives on the gold leaf as they will now arrive at the same answer for the gold leaf.

I do not mean to be a trouble maker here but there is one fly left in the soup. Replace the spring gold leaf with a Faraday cage that has 0 charge. We place one electron in the middle somewhere. As the cage has 0 charge then Q X Q / R^2 or 0 X 1 electron = 0 force. The electron is free to float around not effected by a Coulomb force.  I am going to throw a wrench in the works by sprinkling some electrons on the Faraday cage. let us say the cage now has a charge of -1 and the electron floating inside has a charge of -1 just to give it a number. If Q X Q / R^2 is true then the electron should stay in and exact center of the cage as R^2 means closer to any side of the cage will increase the Coulomb force repelling it. The problem is the electron could care less and will still just float around indifferent to R^2 Coulomb force from the sides of the Faraday cage. This was demonstrated in the video by Professor Eric Rogers ironically using the inverse law R^2 to explain it. Maybe I misunderstood it.

He hints that it is similar to gravity at the center of the world where one weighs 0 pounds. That can be understood. However if the earth were a egg shell and we were inside the egg shell gravity would not pull one towards the shell just like the 1 electron floating around in a positive charged Faraday cage for this example would not feel a Coulomb force to the cage walls . How can an electron not want to go to a positive charged Faraday wall? In an oscilloscope the electron will go toward a positive deflection plate but not in the Faraday cage despite R^2.

I think we are getting close to complete resolution. The question is, if you charge up a Faraday cage, meaning the cage has an excess charge, then why does an electron inside see no force, at ANY position within the cage. Professor Rodgers explained this, but you may have missed his point.

Assume for simplicity that the cage is spherical. At the centre of the cage, we know from symmetry there can be no force on the electron. You might think intuitively that if you move the electron off centre and closer to the inside wall, the net Coulomb force would increase, as it gets closer to the charged wall, especially as the force goes as 1/R^2.

The trick is to realize that the electron is being pulled literally in EVERY direction, and these forces always cancel just as long as the inverse square law is true. Look again at the video. If you move the electron off centre, then you still have to consider ALL the Coulomb forces, not just the forces from the part of the wall nearest to the electron, which is what you are doing. If you cleverly divide the inside surface of the sphere into little ‘patches’, you find that the ‘patches’ on the wall where the wall is nearest to the electron have less area, in inverse proportion to R^2, compared to the larger ‘anti-patches’ on the part of the wall that is further away. The force from any such patch is proportional to it’s area, and therefore inversely proportional to it’s distance R from the electron. Each patch area scales as R^2, but the Coulomb force scales inversely as 1/R^2, with the result that the magnitude of the Coulomb force is the same for every patch, but the forces cancel out because each pair of patches pull in an opposite direction. It’s messy to explain verbally, but if you look again at the video then I’m sure it will be clear, not to mention rather nifty.

You can reach the same conclusion trivially easily by application of Gauss’s Law to an imaginary Gaussian closed surface, chosen to be a concentric sphere inside the spherical Faraday shell. As there is no charge enclosed in the spherical Gaussian surface, the electric field must be zero everywhere on this surface, and thus everywhere inside the spherical Faraday shell. If there is no electric field, then there is no force on an electron, as it is an electric field (units Newtons/Coulomb) that provides electrostatic force on charges. However, this explanation is cheating because you have to take Gauss’s Law on trust, whereas Rodger’s explanation in the video is intuitive and understandable by anyone. Although spherical Faraday cages are easiest to analyse, in fact the electric field is zero everywhere inside ANY closed, conducting shape. For a sphere, the charge excess charge distributes itself evenly on the surface, as it must from symmetry considerations. For other shapes, the surface charge density is not constant, but distributes itself in such a way that the E-field is everywhere zero inside, and it turns out that in order to do so, the charge density is much higher at corners and edges.

Yet another way we know that the E-field must be zero inside a hollow conductor, is to note that the entire conductor must be equipotential, at the same voltage. This we know from Ohm’s Law. At DC, any two point on a conductor are at the same voltage, unless there is a current flowing, which there is not in an electrostatic problem. If every point on the inside surface is at the same voltage, then intuitively one can see that there can be no electric field created, units of V/m. You can’t produce volts per meter, unless you have voltage difference(s), and there can be no voltage differences anywhere on a closed equipotential surface.

Quicky getting back to your intuitive feeling that the electron would see a stronger Coulomb force when nearer to the inside wall, consider the case of an electron above a large, charged flat plate. You may intuitively think that Coulomb force of attraction to (or repulsion from) the plate would be stronger at closer distance to the plate, yet in fact the force is a constant, independent of distance, because as the electron moves further from the plate (force dropping off as R^2), it ‘sees’ more of the plate, increasing by R^2.

Enough raving. I enjoy solving mysteries and paradoxes, or at least trying to, so keep them coming.
 


Share me

Digg  Facebook  SlashDot  Delicious  Technorati  Twitter  Google  Yahoo
Smf