Electric field inside a hollow conductive sphere

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The discussion centers on the behavior of electric fields within a hollow conductive sphere, specifically when a very small charge, such as two electrons, is placed on it. While it is established that the electric field inside a conducting sphere is zero, participants clarify that the presence of a few electrons does not significantly disrupt this condition due to the large number of free electrons in the conductor that can realign themselves. Additionally, the conversation touches on the nature of free electrons in metallic conductors, explaining that these are not extra electrons but rather a subset of the total electrons, with the majority being bound to the atomic structure. The distinction between conducting and valence electrons is made, emphasizing that the overall charge of metals remains neutral despite the presence of free electrons. The thread concludes with insights into the fundamental properties of conductors and their electron behavior.
TonyZ
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Hello,
This is my first post to these forums.

It is well known that the electric field inside a conducting
sphere is zero everywhere inside. But what if the charge
placed on the sphere is very, very small. Let's say it
consists of only two electrons. Clearly, this small charge cannot
be evenly distributed on the surface of the sphere. The
two electrons will undoubtedly end up as far apart as possible,
i.e. at opposite sides of the sphere. It doesn't appear that
the electric field is zero everywhere inside this sphere.

I'd like to get the group's input on this.

Thanks!
Tony
 
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A conducting sphere will have many free electrons, and it will be no problem for them to spread evenly.
A charged sphere of copper might consist of 10^24 copper ions and 10^24+2 electrons in the conduction band that can all move around.
 
The theorems about conductors generally assume a huge number of conduction electrons as willem2 says. If only two electrons are added to a conductor, the lare number of conduction electrons will realign to produce zero field inside. That is, adding two newborns to the world population does not change much, except for their families.
 
Thanks to all who replied! I understand what you are saying.

I have one other question: Are the "free electrons" in a metalic conductor those which are above and beyond the ones that are part of the metal atoms? For example, copper atoms have 29 electrons and 29 protons. Does having free electrons mean there are more electrons than protons? If so, wouldn't that mean that every metal is inherently
negatively charged (which I know is not correct)?

Thanks,
Tony


Tony
 
TonyZ said:
I have one other question: Are the "free electrons" in a metalic conductor those which are above and beyond the ones that are part of the metal atoms? For example, copper atoms have 29 electrons and 29 protons. Does having free electrons mean there are more electrons than protons?

There are as many electrons as there are protons. A fraction of the 29 are conducting and the rest are not (valence electrons). I don't know how many of the 29 are valence and how many are conducting, but it should be a well-known constant number at room temperature for copper specialists (probably in a table somewhere, and quite readily calculable with the basic theory).

In condensed matter/solid state theory, a crystal (including a metallic polycrystal) is considered a lattice of (+) ions, through which conducting electrons flow, not a lattice of (neutral) atoms.
 
I'm working through something and want to make sure I understand the physics. In a system with three wave components at 120° phase separation, the total energy calculation depends on how we treat them: If coherent (add amplitudes first, then square): E = (A₁ + A₂ + A₃)² = 0 If independent (square each, then add): E = A₁² + A₂² + A₃² = 3/2 = constant In three-phase electrical systems, we treat the phases as independent — total power is sum of individual powers. In light interference...

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