B Question about understanding conductors for EM course

[Dale]

I wouldn't say it is misleading. It is just unfamiliar to you and so you have not yet built up a correct intuition.

 

[Guest812]

Yes, I understand this already; in post 42 and 44 I was referring to a circular conductor. (a sphere)

If the sphere diameter is infinitely large, near the infinitely large sphere's surface, the sphere's E-fields will behave as if the sphere's surface was flat.

A tiny sphere, can treated as an infinitely large sphere, if you measure extremely close to the sphere's surface. If you know how the E-fields behave extremely close to the sphere's surface, then by using symmetry and logic, you can probably determine how they behave further away on both sides of the surface.

Or in other words, a sphere (regardless of diameter) having excess surface charge, will have radially directed E-fields, directed both outward away from the center and inward toward the center all around the sphere. The inwardly directed E-fields will totally cancel each other everywhere within the sphere, but add to the outwardly directed E-fields on the opposite side.

 

[Sho Kano]

but add to the outwardly directed E-fields on the opposite side.

I didn't know about this, but it makes sense because a charged particle would experience a force on the other side due to THAT surface charge.

The inwardly directed E-fields will totally cancel out each other everywhere within the sphere

Yes, so if I place a charged particle at a radius (lets say the radius of the sphere is 1 meter) 0.75 meters from the center of the sphere, it would experience no force. So there is a crazy thing going on that the surface charges don't contribute at all inside the conductor.

 

[Guest812]

...

Yes, so if I place a charged particle at a radius (lets say the radius of the sphere is 1 meter) 0.75 meters from the center of the sphere, it would experience no force. So there is a crazy thing going on that the surface charges don't contribute at all inside the conductor.

Yes.

In my above posts, I diagrammatically explained it using a thick flat conducting plate with equal surface charge on both opposite sides analogy, then explained how a conducting sphere of infinite diameter could be treated as a conducting flat plate, and how even the tiniest conducting sphere when "viewed" extremely close to its surface can be treated as a sphere of infinite diameter. No matter what shape the surface is, when viewed extremely close to its surface, that infinitesimally small surface area will appear as an infinitesimally small flat plate surface area.

The best "no E-field anywhere inside a conducting sphere" explanation involves diagrams and calculus. The easiest mathematical approach I've seen involves Guass' Law, and the "Solid Angle" concept. But as I explained above, before attempting this more mathematically challenging spherical problem, it would probably be wiser (easier) to mathematically solve first the infinitely thin single charged plane, and then solve the two parallel infinitely thin surfaces charges separated by a thick long flat conducting plate. All the decent college calculus based Physics books explain those flat plane calculations, and the better Physics books will mathematically derive why the E-field is zero everywhere inside a conducting sphere, as well as why the E-field is zero anywhere inside any shaped conductor.

 

[Sho Kano]

Or for other words, a sphere (regardless of diameter) having excess surface charge, will have radially directed E-fields, directed both outward away from the center and inward toward the center all around the sphere.

Right, a flat conducting plane will have a constant field as long as you are close to it. So now, since any sphere can be treated as a plane (close up), and planes have a constant field, it wouldn't matter if you were .75 from the center. Is this what you're getting at? Again, this is very non-intuitive.

 

[Guest812]

Right, a flat conducting plane will have a constant field as long as you are close to it. So now, since any sphere can be treated as a plane (close up), and planes have a constant field, it wouldn't matter if you were .75 from the center. Is this what you're getting at? Again, this is very non-intuitive.

The text I change to red is incorrect, in the context you used it in. The correct and more accurate phrasing is:

"an infinitely long flat conducting plane will have a constant E-field, regardless of how far away from the plane the E-field is measured.

Note an infinitely long flat conducting plane will look exactly the same, regardless of how close you are to it. If you're an inch away from an infinitely long surface, take a photo, then magnify the photo so an inch distance looks like a mile, the plane will still look infinitely long, but now you look like you're a mile away from the plane.

Like others and I have been repeatedly trying to explain to you, before you think about the E-field inside a sphere, it would be easier if you first understand the E-field on both sides of a long infinitely thin conducting surface charge, then think about the E-field on both sides of a long thick conducting surface with surface charges on both sides. These examples are standard beginner college level Physics E&M problems. If you can't solve these beginner type problems, you'll probably be less able to solve more mathematically advance problems like calculating the E-field inside a conducting sphere.

 

[Sho Kano]

The text I change to red is incorrect, in the context you used it in. The correct phrasing is:

"a long flat conducting plane will have a constant field, regardless of how away from it you are.

Only in the case of an infinitely large plane. There are none, and I can't see how that logic can be applied to a non-infinite, limited sphere of radius r.

before you think about the E-field inside a sphere, it would be easier if you first understand the E-field on both sides of a long infinitely thin conducting surface charge,

Like I said, I already understand that. The field is given by \frac { \sigma }{ 2{ \varepsilon }_{ 0 } }

then think about the E-field on both sides of a long thick conducting surface with surface charges on both sides.

Putting the two parallel to each other (assume negative charges on both), just superposition the fields. On top of the plate is just \frac { \sigma }{ { \varepsilon }_{ 0 } }In the middle is 0. Below is again, \frac { \sigma }{ { \varepsilon }_{ 0 } }.

 

[Guest812]

Only in the case of an infinitely large plane. There are none.

Like I said, I already understand that as I already completed my EM course. The field is given by \frac { \sigma }{ 2{ \varepsilon }_{ 0 } }

Putting the two parallel to each other (assume negative charges on both), just superposition the fields. On top of the plate is just \frac { \sigma }{ { \varepsilon }_{ 0 } }In the middle is 0. Below is again, \frac { \sigma }{ { \varepsilon }_{ 0 } }.

Good.

Then from there, the only way I know from the top of my head to intuitively view what happens inside and outside a conducting sphere, is visualize how the E-field behaves when it is measured infinitely close to both sides of the sphere's surface, where infinitely close to a sphere's surface means so close that the sphere appears to be flat. Like the way the Earth appears when you stand on Earth, you're so close to the Earth's round surface that for all practical purposes the Earth's round surface appears flat to you.

Chapter 4 of this free downloadable E&M text from MIT explains how to calculate and visualize E-fields inside and outside conductors very well: http://ocw.mit.edu/courses/physics/...ng-2007/class-activities/chapte4gauss_law.pdf

Also the free Video Lecture from Yale's Physics II 201 course taught by professor Shankar has excellent explanations and mathematical derivations: http://oyc.yale.edu/physics/phys-201/lecture-4

 

[Sho Kano]

visualize how the E-field behaves when it is measured the infinitely close to both sides of the sphere's surface, where infinitely close to a sphere's surface means so close that the sphere appears to be flat.

Yes, like in post 56, the field is constant.

Like you way you feel when you stand on Earth, you're so close to the Earth's round surface that for all practical purposes it appears flat to you.

Like how gravitational field can be approximated to g? What if I go closer to the center, does it still stay constant?

 

[Guest812]

Yes, like in post 56, the field is constant.

Like how gravitational field can be approximated to g? What if I go closer to the center, does it still stay constant?

NO !

A sphere only appears as a flat surface when you're extremely close to the sphere's surface. The further away from the sphere you are, the more the sphere will appear as a point charge.

In other words, when viewed extremely close to the conducting sphere's surface, the E-field lines emanate perpendicular from both externally and internally from a sphere's surface, which when viewed extremely close to the sphere's surface the E-field lines appear to emanate parallel to each other; however, perpendicular to the spheres surface when viewed far away from the surface is actually radial: meaning externally from the sphere the further away from the sphere the weaker the E-field (i.e. externally the radial E-fields aren't constant, instead they weaken as 1/R^2), and internally the closer towards center the stronger the E-field; however for each internal E-field emanating from one side of the sphere there will be an exact opposite inward radially directed E-field emanating from the opposite side of the sphere. Since every point of the sphere generates an inward E-field exactly opposite of the opposite side of the sphere, every E-field is exactly canceled inside the sphere, the same way they did inside an infinitely long thick flat conducting plate.

I have a (bad) habit of editing (almost all my posts) after posting them. I think the best explanations will be found in the last two comments at the end of my earlier post #59 of this thread.

 

[Sho Kano]

I have a (bad) habit of editing (almost all my posts) after posting them. I think the best explanations will be found in the last two comments at the end of my earlier post #59 of this thread.

Thanks for the links. I think the surface charges can simply be ignored because they cancel each other below the surface, similar to when calculating the force of gravity inside earth. The notion of the plate and sphere proves that the field outside is just like one due to a point charge

 

[Dale]

So there is a crazy thing going on that the surface charges don't contribute at all inside the conductor

They do contribute. They contribute to making the field 0. If they didn't contribute then the field would be the same inside as outside the conductor. They contribute so as to make the field 0 inside regardless of the field outside.

 

[Sho Kano]

I'm not so sure the balloon will move if the external field is uniform. In such a field the force on every charged particle should be equal. If the field is stronger on the left side than the right side however, then you'd have a situation where the negative charges have moved into the region where the field is stronger. The attractive force would be stronger on these charges than the repulsive force on the positive charges on the right side of the balloon.

The same is for insulators right? Although it allows electric fields to penetrate through, an insulating balloon shouldn't move in the presence of a uniform E field.

 

[Guest812]

The same is for insulators right? Although it allows electric fields to penetrate through, an insulating balloon shouldn't move in the presence of a uniform E field.

An uncharged insulated balloon (or anything else: even a conductor) will experience an inductive electrostatic force (i.e. if lite enough will accelerate and move) if inserted into any E-field. The reason anything will experience an electrostatic force if inserted into any E-field is because any E-field will pull the object's electrons one way, and push the object's proton's (i.e. nucleus) in the opposite direction. When an object's electrons and protons/nucleus are separated from their atom's center, the atoms will become polarized, and polarized objects are attracted to E-fields because one side of the polarized object will be closer to the E-field than the opposite side. That is why the E-field from a charged insulated comb is able to electrostatic ally attract both insulated and conductive items. The insulated items will remain clung to the comb, but the conductive items will want to share the charge with the comb, then when the conductor becomes charged, the now equally charged conductor will want to repel from the comb.

 

[Dale]

An uncharged insulated balloon (or anything else) will experience a force (i.e. move) if inserted into any E-field.

No, this is incorrect. Sho Kano is correct. There will be no net force on an uncharged object in a uniform E field. There can be a net force if the field is non uniform, and there can be a torque even if the field is uniform, but not a net force on an uncharged object in a uniform field.

 

[Guest812]

No, this is incorrect. Sho Kano is correct. There will be no net force on an uncharged object in a uniform E field. There can be a net force if the field is non uniform, and there can be a torque even if the field is uniform, but not a net force on an uncharged object in a uniform field.

Then please explain how a charged object (i.e. something that produces an E-field) can attract any (i.e. insulating or conductive) uncharged object?

In the meantime, I'll try to find the chapter in the free MIT text that explains the phenomenon better than I did.

 

[Dale]

Then please explain how a charged object (i.e. something that produces an E-field) can attract any (i.e. insulating or conductive) uncharged object?

The field around a charged object is generally not uniform.

 

[Sho Kano]

Then please explain how a charged object (i.e. something that produces an E-field) can attract any (i.e. insulating or conductive) uncharged object?

In the meantime, I'll try to find the chapter in the free MIT text that explains the phenomenon better than I did.

Drakkith discusses this on post 38
My understanding is that an outside field polarizes an insulator or conductor (in the case of an insulator the electrons are still attached to the atoms).
There will then be attractive force on one side, and a repulsive on the other (depending on the direction of the outside field), which cancels each other out ONLY if the outside field is uniform.

 

[Guest812]

OK guys, thanks for setting me straight.

I drew a uniform E-field, then drew a nonconductive object inside the field, then electrostatically induced "polarized" charges on opposite sides of the object, and after looking at the diagram, agree no net movement of the complete object, just electrons pushing just as much towards one side of the insulator as protons are pushing towards the opposite side of the insulator, therefore no net electrostatic force on the insulator in a uniform E-field.

Therefore I can't think of any reason why an uncharged conductor should move any differently than an insulator inside a uniform E-field. In both the insulator and conductor cases, when inserted inside a uniform field, the objects electrons will be separated more towards one side of their positively charged protons. But both the electrons and protons will exert equally oppositely directed forces on both the insulator and conductive object.

 

[Stavros Kiri]

It is only zero in the electrostatic condition. I.e. given that a system is electrostatic then $J=0$. In a conductor $J=\sigma E$ so $E=0$.

I think this is also the principle of 'Electromagnetic shielding', i.e. using hollow [metal] conductors to achieve E = 0 inside ...

 

[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.