Why do electric fields seem to be zero in conductors?

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SUMMARY

The discussion clarifies that electric fields within conductors are zero due to the movement of free electrons in response to external electric fields, achieving electrostatic equilibrium. When an electric field is applied, free electrons redistribute themselves, creating an opposing field that cancels the internal field. This results in a net electric field of zero inside the conductor. The conversation also touches on the limitations of this theory when charge carriers are constrained and the implications of charge distribution in non-classical scenarios.

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  • Familiarity with conductors and charge carriers
  • Knowledge of electrostatic equilibrium principles
  • Basic grasp of classical physics concepts related to charge distribution
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Berko
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Every book seems that it is a no-brainer that electric fields are zero in a conductor, but I do not understand this.

First of all why assume it is zero just because, by contradiction, any electric field will produce acceleration in charges which cannot happen when there is electrostatic equilibrium. Perhaps there is equilibrium because all the charges have moved and there is no free charge that can move.

Second, how exactly does the negative charge go to the one side of the conductor and the positive charge to the other side? The negative charge are the free electrons (which can move) while the positive charge are the lattice ions (which cannot).
 
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Berko said:
Every book seems that it is a no-brainer that electric fields are zero in a conductor, but I do not understand this.

First of all why assume it is zero just because, by contradiction, any electric field will produce acceleration in charges which cannot happen when there is electrostatic equilibrium. Perhaps there is equilibrium because all the charges have moved and there is no free charge that can move.

Second, how exactly does the negative charge go to the one side of the conductor and the positive charge to the other side? The negative charge are the free electrons (which can move) while the positive charge are the lattice ions (which cannot).

At the instant the conductor is in an electric field, there is a net field inside. However, the electrons will move to their respective side and we can approximate the opposite side as mostly positive. The electrons will distribute in such a way that if we were to now ignore the external electric field and just regard the conductor now as a dipole, the electric field vectors will be equal in magnitude but in opposite direction as that of the external field. Applying superposition principle now, the net field inside is zero, and the system is NOW in electrostatic equilibrium.

Hope this helps
 
Or in another light, electrons are free to move inside the conductor. Suppose we have an electric field exist inside a conductor, electrons inside the conductor will immediate move in the direction according to the polarity of the field. When electrons move towards the field, it will generate an opposite field. Electrons will keep move until every bit of the field inside the conductor is canceled out. So there will be no field inside a conductor.

Just another way to say the same thing.
 
Berko,

I like your question very much, even though it is very unrealistic.
Many other phenoma could could come into play before this limitation should be considered.
For example, charges can leave a conductor, if the field is high enough.

You could evaluate a typical density of available charge carriers in a piece of metal, and calculate from this the external field that would move all these charge and bring you to the limit os the standard theory for conductors. The standard theory of conductors, assumes (correctly) that the available charge has no limit.

It could also be interresting to try to calculate the charge distribution and the field penetration when the amount of charge carriers is limited. It might even not be so obvious to translate this problem in clear mathematical terms. It would imply making a difference between the positively-charged parts and the negaively-charged part on the surface of the conductor, and putting then a constraint on the total charge on the positive (or negative) side. The maximum charge would then also depend on the bulk volume of the conductor. I guess that the charge would still accumulate on the surface, but this should be checked and proved again.

I like this problem because I am curious to know if it could be translated into an optimization problem. I guess that electrostatics can be derived from a variational principle, even when the amount of charge carriers is constrained.
 
Mr. Berko I'll answer the second part of your question.
Yes the electron don't toally go and get collected on one side( when in electric field) as the whloe distribution will become absurd (as you've rightly stated that the lattice ions can't drift but this wasn't known at the time of Coulomb thus this approximation was made).
So it is like this - when in electric field the electron distribute themselves on the conductor such that the field isn't affected by its presence, the distribution is such that we would apparently have -ve charge on one side and +ve on another (but not in real sense).
But if you want to calculate the distribution of electrons on the conductor kept in an electric field then it'll go into non-classical realms
 
Deric Boyle said:
Mr. Berko I'll answer the second part of your question.
Yes the electron don't toally go and get collected on one side( when in electric field) as the whloe distribution will become absurd (as you've rightly stated that the lattice ions can't drift but this wasn't known at the time of Coulomb thus this approximation was made).
So it is like this - when in electric field the electron distribute themselves on the conductor such that the field isn't affected by its presence, the distribution is such that we would apparently have -ve charge on one side and +ve on another (but not in real sense).
But if you want to calculate the distribution of electrons on the conductor kept in an electric field then it'll go into non-classical realms

I once did a back of the envelope calculation in a thread a while back concerning this. What you can do is assume that you have a rectangular block of copper and move a single electron from the lattice atoms on one face and move it to the other side of the block. The resulting electric field is astronomical. The size of the electric field that would have to be applied to a good conductor to strip off just the surface electrons is far greater than any fields we work with. So in the end, it is a very good approximation and it is a simple exercise to prove this to yourself.
 
Thank you all!
 

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