Question about understanding conductors for EM course

[Sho Kano]

I know this a long thread, but bear with me, I still have some uncertainties
-Say the same situation with a neutral conductor in a E field, what if the field outside was increased so that there are not enough electrons/protons to cancel it out? Does this ever happen?
-It seems like in the situation that we have charged balloon, it would move even in the presence of a uniform E field, because there would be uneven amounts of charge on each side.
-If the first situation was true (field lines do end up penetrating), there would still be no movement of charge because the charges are already at the boundaries of the conductor (at the surface on each side)- doesn't that violate J=sigma*E?

 

[Stavros Kiri]

I know this a long thread, but bear with me, I still have some uncertainties
-Say the same situation with a neutral conductor in a E field, what if the field outside was increased so that there are not enough electrons/protons to cancel it out? Does this ever happen?
-It seems like in the situation that we have charged balloon, it would move even in the presence of a uniform E field, because there would be uneven amounts of charge on each side.
-If the first situation was true (field lines do end up penetrating), there would still be no movement of charge because the charges are already at the boundaries of the conductor (at the surface on each side)- doesn't that violate J=sigma*E?

I think that Dale is right in having said that such intuitions may be misleading. All it matters is that the equations are met with correct calculations. Everything should be consistent. However I will attempt to justify the situation for the 3 cases that you are describing above:

1.&3. J = σ E for a conductor, always. In electrostatic equilibrium it has to be J = 0. Thus E = 0 inside the conductor, no matter what. Even for a very strong external field, there has to be some charge distribution inside the conductor that brings E = 0, or otherwise it's not the equilibrium yet. Just think of it as the appropriate distribution, where the charge density (as a positional function) will be eventually solved and given in terms of E(external), and it will be a function of space. But it will have a defined value, even for a very strong field. But the key here is I think 'charge density', and not total or partial charge (which could perhaps have a limit). Just recall the diffetential form of Gauss's law. [divE = ρ(x,y,z)/ε₀, where ρ(x,y,z) the charge density as a function of space and divE = ∇E, etc. ]
Just allow for the math to be hypothetically carried out and you will not have any more questions. Electromagnetism is a beatiful and consistent theory.

2. Not if the total force is zero. Is it?

 

[Dale]

ike in the situation that we have charged balloon, it would move even in the presence of a uniform E field

Yes. Although I would say "experience a net force from the E field" rather than "move" since the actual motion depends on the initial velocity and any other forces that may be acting.

 

[Stavros Kiri]

-It seems like in the situation that we have charged balloon, it would move even in the presence of a uniform E field, because there would be uneven amounts of charge on each side.

+ I would say better because F = Q E, thus there is a net force due to the toral charge Q and the uniform E field, not because of uneven distribution ...

 

[Stavros Kiri]

+ I would say better because F = Q E, thus there is a net force due to the toral charge Q and the uniform E field, not because of uneven distribution ...

Unless you see it from the point of view of the sum of all force components (or better integral), roughly F_tot = ∑_i F_i ... , on all charge "parts" ...

Intuition is good, but math works better and is less misleading. (We just have to be careful to do both right.)

 

[Sho Kano]

+ I would say better because F = Q E, thus there is a net force due to the toral charge Q and the uniform E field, not because of uneven distribution ...

IS there an uneven charge distribution?
Since there is say more electrons in the side where the field is, the attractive force from that side would exceed the repulsive one on the other. So the balloon moves or experiences a net force.

 

[Stavros Kiri]

IS there an uneven charge distribution?
Since there is say more electrons in the side where the field is, the attractive force from that side would exceed the repulsive one on the other. So the balloon moves or experiences a net force.

Correct! There is an uneven distribution. Both ways are consistent (+ look my other comment - "reply" to myself) if you sum the forces, given that E = const (uniform field).

E.g. for the two sides (easy math): F_tot = F1+F2 = q1 E + q2 E = (q1+q2) E = Q E, but careful to put the signs (poss. neg. for charges if ..., + repulsive, attractive forces etc. ...).

 

[Drakkith]

-Say the same situation with a neutral conductor in a E field, what if the field outside was increased so that there are not enough electrons/protons to cancel it out? Does this ever happen?

That's one hell of a field. I don't know for sure, but I think that you'd start ripping electrons off of the metal's surface before that would happen.

-It seems like in the situation that we have charged balloon, it would move even in the presence of a uniform E field, because there would be uneven amounts of charge on each side.

Do the math for yourself with a couple of charges on each side of a conductor in a uniform electric field.

 

[Stavros Kiri]

However, sometimes math in E&M is not that easy (e.g. uses vector calculus and differential geometry) [cf. my other comment-reply to yours, with the divE and ∇E symbols]. That happens especially with perhaps more complicated non-uniform charge distributions. But do not be scared. Even then we can use theorems (e.g. Gauss's theorem, Stokes' theorem) or e.g. the integral form of Gauss's law of E&M, or other tricks (such as the "image method", etc.), that make things a lot easier.

Some of your questions can and have been intuitively attacked and answered. Intuition and picturing things is also useful. But in hard or ambiguous cases the math is the safest way, and sometimes even the only one.

 

[Stavros Kiri]

That's one hell of a field. I don't know for sure, but I think that you'd start ripping electrons off of the metal's surface before that would happen.

It depends which one is the fastest: A) electrons to be ripped off [you need to break and destroy e.g. the metal bonds, which are pretty tough!?] or B) reach of equilibrium?[that happens very fast e.g. in metals (due to the free electrons), that have a fast high, almost immediate response to compansate for field changes (e.g. see RF shielding ...)].

I vote for A), in most cases, but anything is possible. What do you think?
In other words can we say that in most known cases E= 0 inside a conductor? (in fast equilibrium, because J=0 ...)

 

[Drakkith]

A) electrons to be ripped off [you need to break and destroy e.g. the metal bonds, which are pretty tough!?]

I think you would just need to exceed the work function of the metal, but that's mostly a guess. You'd only need to break metallic bonds if you were removing whole atoms.

 

[Stavros Kiri]

I think you would just need to exceed the work function of the metal, but that's mostly a guess. You'd only need to break metallic bonds if you were removing whole atoms.

You are probably right.

 

[Stavros Kiri]

Here is another question for the OP or others, which is, I think, crucial in learning about conductors:

Assume a closed surface hollow metal conductor. In equilibrium E = 0 inside the metal. What is the field inside the empty space in the hollow area and why?
[Please prove or justify fully your answer for it to be accepted. Let us give priority to the OP first.]

 

[Sho Kano]

Here is another question for the OP or others, which is, I think, crucial in learning about conductors:

Assume a closed surface hollow metal conductor. In equilibrium E = 0 inside the metal. What is the field inside the empty space in the hollow area and why?
[Please prove or justify fully your answer for it to be accepted. Let us give priority to the OP first.]

In a closed surface (say spherical) hollow conductor at equilibrium (meaning uniform distribution), there is no field inside the empty space. This is because the E field from the surface charges cancels everywhere inside- this is true even if you're not in the center. I don't really know the specifics, but I have attached a (pretty bad) picture of how I currently think it goes. The contributions from the top of the line will exactly cancel with the contributions on the bottom (less charges, but more intense on top. more charges, but less intense on the bottom). I'm probably incorrect.

 

[Stavros Kiri]

In a closed surface (say spherical) hollow conductor at equilibrium (meaning uniform distribution), there is no field inside the empty space. This is because the E field from the surface charges cancels everywhere inside- this is true even if you're not in the center. I don't really know the specifics, but I have attached a (pretty bad) picture of how I currently think it goes. The contributions from the top of the line will exactly cancel with the contributions on the bottom (less charges, but more intense on top. more charges, but less intense on the bottom). I'm probably incorrect.

You have a good intuition and I think it is correct here (from line to top less charges but stronger field, while from line to bottom more charges but weaker field .../ and you do in fact get zero if you actualy do the integral for any point inside - in the uniform symmetrical case).

And a similar intuition can be built for non-uniform distribution (of charges) and arbitrary [closed] shape hollow conductor, at equilibrium. In fact this holds even in the presence of an arbitrary external field E (and in fact that's where it gets more interesting). It also constitutes the basis principle of Electromagnetic Shielding (RF shielding is a special case [RF = Radio Frequency]) [cf. Wiki..]. Usually metal is preferred. Such cases of conductors are also known particullarly in the Electrostatic case as "Faraday cage".

The intuition behind it (also along the lines of Sho Kano) is as follows: In the equilibrium (i.e. with [and/or constrained by] some time response parameters, e.g. appropriate frequences vs restoration times [in the case of time varying fields]) the charges inside the conductor will necessarilly arrange themselves in such a way so that the total field is zero on the conductor and inside its hollow.

But why does this hold? What is the mathematical proof (of that theorem)? Anyone is welcome now.

Note: for inside the conductor, the proof was given earlier by Dale (J = σE , thus for J = 0 we get E = 0). Therefore only the proof for the hollow area remains.