The shell theorem and particles

[FS98]

As two particles become closer to each other, the gravitational force (or electric force) approaches infinity. If this is the case, then how does the Shell theorem work?

If two particles are extremely close together, there should be an extremely large force. If we then build a sphere around one of the particles, according to the Shell theorem, the force should be the same as in a situation where all of the mass or charge were at the center of the sphere. If we could treat the new sphere like all of the mass or charge were in the center, the force would be finite because there is a finite radius.

How does this make sense? The particle that existed before adding the sphere applied an infinite force on the the other particle, and the addition of the sphere should only increase the force upon the other particle. Instead it seems like the force would decrease. Where am I going wrong?

 

[davenn]

As two particles become closer to each other, the gravitational force (or electric force) approaches infinity.

why do you say that ?

If two particles are extremely close together, there should be an extremely large force.

again, why ?
you are dealing with a finite masses and/or charges, so the gravitational force or electric fields must also have a finite value

go have a read of Newton's law of universal gravitation Dave

 

[FS98]

why do you say that ?
again, why ?
you are dealing with a finite masses and/or charges, so the gravitational force or electric fields must also have a finite value

go have a read of Newton's law of universal gravitation Dave

Gravitational forces and electric forces follow inverse square laws do they not? So as r approaches 0, shouldn’t force approach infinity?

Gm1m2/r2 regardless of the masses seems like it approaches infinity as r approaches 0.

 

[davenn]

Gravitational forces and electric forces follow inverse square laws do they not?

yes but let's take an example of 2 electrons ( same negative charge) so they are repulsive. This is OPPOSITE to the gravitational force of attraction.
So there is some cancelling of the forces

scroll down this page for some worked examples (no point me typing it all out)

https://en.wikibooks.org/wiki/FHSST_Physics/Electrostatics/Electrostatic_Force

Gm1m2/r2 regardless of the masses seems like it approaches infinity as r approaches 0.

two golf balls touching ... is the force of gravity between them infinite ?
you and the Earth touching ... is the force of gravity between you and it infinite ?
Dave

 

[davenn]

in that link I gave, pay attention particularly to
Worked Example 60 Coulomb's Law II[edit]
Question: Determine the electrostatic force and gravitational force between two electrons 1Aring apart (i.e. the forces felt inside an atom)

note the final results in the difference between the gravitational attraction and the electrostatic repulsion.
Two electrons cannot touch ( have zero distance between them) under normal conditions because the electrostatic repulsion is many magnitudes greater than the gravitational attraction between them.

When particle physicists (I'm not one) collide particles together, they must add lots of energy to the system to overcome these forces

Where's @ZapperZ when you need him

Dave

 

[FS98]

yes but let's take an example of 2 electrons ( same negative charge) so they are repulsive. This is OPPOSITE to the gravitational force of attraction.
So there is some cancelling of the forces

scroll down this page for some worked examples (no point me typing it all out)

https://en.wikibooks.org/wiki/FHSST_Physics/Electrostatics/Electrostatic_Force

two golf balls touching ... is the force of gravity between them infinite ?
you and the Earth touching ... is the force of gravity between you and it infinite ?
Dave

If we have a situation where something like 2 golf balls are touching, aren’t there just other forces at play? If golf balls only experienced the electric force, I would think that the force would approach infinity as they became closer and closer together.

If we two imaginary particles that only had charge and no mass were something like 1/10^100000000000m away from each other there would be an enormous force between them. If one of those two particles were surrounded by other particles that formed a sphere of a radius of 5m, I don’t see how the Shell theorem could apply. A sphere 5m away should apply a much smaller force than a particle 1/10^100000000000m away, even with the additional charges. And if we’re discussing hypothetical particles with only charge, there is no canceling going on.

 

[davenn]

If we have a situation where something like 2 golf balls are touching, aren’t there just other forces at play?

what other forces ?

If golf balls only experienced the electric force, I would think that the force would approach infinity as they became closer and closer together.

opposite force or same forces ?

Did you read and get a basic understanding in those examples on that www site ?
maybe not

If we two imaginary particles that only had charge and no mass were something like 1/10^100000000000m away from each other there would be an enormous force between them.

you cannot have imaginary ... imaginary doesn't exist ... stick with real situations ... again, look at those examples
keep it real, stick to science

 

[A.T.]

If golf balls only experienced the electric force, I would think that the force would approach infinity as they became closer and closer together.

The "r" in the inverse square law is the distance between the centers, not between the surfaces.

 

[FS98]

The "r" in the inverse square law is the distance between the centers, not between the surfaces.

But that’s because of the shell theorem isn’t, which is what I don’t understand. It seems like there should be 2 points on the golf balls where the radius was 0 and force was infinite (if by touching we mean that there is some point with no distance between the two objects, which I know wouldn’t happen). In this situation the force seems like it would have to be infinite, but the Shell theorem seems like it says that it will actually be much smaller.

 

[DrStupid]

It seems like there should be 2 points on the golf balls where the radius was 0 and force was infinite (if by touching we mean that there is some point with no distance between the two objects, which I know wouldn’t happen).

But the two points have no volume and therefore no mass (as long as the density is finite).

 

[ZapperZ]

Where's @ZapperZ when you need him

Dave

You didn't use the Bat Signal! :)

I read, and reread through this thread, and I STILL don't quite get what the problem is.

The so-called "shell theorem" is a direct result of a Gauss's Law-type description for electric field and gravitational field. So let's get that out of the way first.

Now, from what I can barely decipher, the OP is asking about the situation where, if we have a finite-sized sphere, say radius R, having some mass or some charge, and then another mass or charge touches the surface, why isn't the force due to the relevant field infinitely big since two charges or two masses are in very close proximity to one another?

Is this about right?

If it is, then this is the case where you are using one set of description and applying it to a situation where it doesn't apply.

The shell theorem/Gauss's law for a non-point object ASSUMES that you have a CONTINUOUS distribution of the source (charge or mass). Once you do that, then the resulting field is the sum of the fields from the entire continuous charge distribution. There are no longer "point charges" anywhere. The distance that matters now is not the distances between each "dm" or "dq" in the mass or charge distribution, but rather the distance from the center of the sphere (we're still working with the highly symmetric situation with the sphere of radius R).

Now, if you start delving into the issue that the sphere itself is made up of these "atoms" and charges and they each are individual point charges, then you no longer have a continuous charge distribution, but rather a granular picture of the distribution. Now, you no longer are using classical E&M to describe the situation, but rather quantum mechanics to figure out how a point charge interact with another point charge that is part of a solid. In other words, the rules of the game is now different! An electron approaching the surface of a spherical conductor versus an electron approaching the surface of a spherical insulator will have different sets of rules. And if you zoom in even more, the complicated issues of surface bonds, vacuum-solid interface, image charge, electron affinity, will all come into play even before we consider the electric force. It isn't trivial.

This is where the macroscopic picture makes it easy, and also why the macroscopic picture can't be used when you decide to dig under the covers.

Zz.

 

[FS98]

How about this situation?

We have a spherical shell with a uniform charge distribution.

The electric field anywhere on the inside should be 0.

If we treat the shell as if all of the charge were at its center, wouldn’t there be an electric field at any point inside the sphere?

 

[ZapperZ]

How about this situation?

We have a spherical shell with a uniform charge distribution.

The electric field anywhere on the inside should be 0.

If we treat the shell as if all of the charge were at its center, wouldn’t there be an electric field at any point inside the sphere?

If you do that, then you're applying the physics all wrong!

The Gauss's law description is different for the outside solution and the inside solution. You are welcome to look up the inside solution, which actually will give you ZERO as the field inside the shell. The outside solution having the 1/r² solution works ONLY for outside the shell.

Zz.

 

[FS98]

If you do that, then you're applying the physics all wrong!

The Gauss's law description is different for the outside solution and the inside solution. You are welcome to look up the inside solution, which actually will give you ZERO as the field inside the shell. The outside solution having the 1/r² solution works ONLY for outside the shell.

Zz.

I know how to apply gauss’s law to to a spherical Shell and get 0 for the inside. I still don’t see how that is the same as if the charge is in the center though. If that charge is q, then why wouldn’t the electric field surrounding that charge be E = kq/r^2? That is the equation for an electric field from a point particle is it not?

 

[ZapperZ]

I know how to apply gauss’s law to to a spherical Shell and get 0 for the inside. I still don’t see how that is the same as if the charge is in the center though. If that charge is q, then why wouldn’t the electric field surrounding that charge be E = kq/r^2? That is the equation for an electric field from a point particle is it not?

I don't understand your question.

You stated that you have a spherical shell of charge, and that the outside solution looks like all the charges are at r=0. Fine.

But then you went INSIDE the shell. There is no longer any charge at r=0! The solution for inside the shell does show that. There is no longer a 1/r² dependence of E!

http://physicstasks.eu/1531/field-of-charged-spherical-shell

So what are we talking about here?

Zz.

 

[FS98]

But then you went INSIDE the shell. There is no longer any charge at r=0! The solution for inside the shell does show that. There is no longer a 1/r² dependence of E!

Zz.

But if we use the shell theorem, can’t we say that all of the charge exists at r=0 because that is the center of a uniform sphere. Then I no longer see how you can get a solution of 0.

I understand how to get the 0 if we don’t pretend that all of the charge is at the center. I don’t understand how it can be 0 if we do pretend all of the charge is at the center.

 

[ZapperZ]

But if we use the shell theorem, can’t we say that all of the charge exists at r=0 because that is the center of a uniform sphere. Then I no longer see how you can get a solution of 0.

I understand how to get the 0 if we don’t pretend that all of the charge is at the center. I don’t understand how it can be 0 if we do pretend all of the charge is at the center.

Again, I have no idea what you are saying here. Have you looked at the link that I gave you that derived the electric field for ALL regions of space for the shell? Look at it, and tell me what EXACTLY it is that you do not understand.

Zz.

 

[FS98]

Again, I have no idea what you are saying here. Have you looked at the link that I gave you that derived the electric field for ALL regions of space for the shell? Look at it, and tell me what EXACTLY it is that you do not understand.

Zz.

The link seemed to explain how to use Gauss’s law to find the electric field in a hollow sphere. I understand that part.

If we have a hollow sphere with total charge q, the field inside is 0, everything is fine.

Now what is the electric field, where the hollow sphere was, of a single charged particle at the location where the center of the sphere used to be?

When I apply gauss’s to this new situation I no longer get 0.

Does the shell theorem not imply that you can treat a spherically symmetric object as if all of the mass or charge is located at the center?

Or did I do something wrong in finding the electric field from the point particle that exists at the center of the shell?

 

[ZapperZ]

Now what is the electric field, where the hollow sphere was, of a single charged particle at the location where the center of the sphere used to be?

When I apply gauss’s to this new situation I no longer get 0.

Does the shell theorem not imply that you can treat a spherically symmetric object as if all of the mass or charge is located at the center?

Or did I do something wrong in finding the electric field from the point particle that exists at the center of the shell?

There have been so many turns in this thread, I have gotten lost many times.

So now it is a DIFFERENT problem. I thought we ALL know the E-field for a point charge? So if you remove the shell, and put a point charge at r=0, then you get the familiar 1/r² dependence once again. So what is the issue now?

Zz.

 

[FS98]

There have been so many turns in this thread, I have gotten lost many times.

So now it is a DIFFERENT problem. I thought we ALL know the E-field for a point charge? So if you remove the shell, and put a point charge at r=0, then you get the familiar 1/r² dependence once again. So what is the issue now?

Zz.

My issue is that I’m getting different electric fields for point charges, and for hollow spheres with the same total charge.

The shell theorem seems to suggest that they should be the same.

 

[ZapperZ]

My issue is that I’m getting different electric fields for point charges, and for hollow spheres with the same total charge.

The shell theorem seems to suggest that they should be the same.

It should not take that much effort to search for the solution to the Gauss's law problem for both situation. And there is no way for anyone to diagnose the problem simply from the way you've described it.

Since this is now a schoolwork-type problem, you should present this in the HW/Coursework forum.

Zz.

 

[jbriggs444]

The shell theorem seems to suggest that they should be the same.

The shell theorem flat out states that they are the same -- in the region outside the shell. A correct statement of the theorem distinguishes between the situation inside and outside the shell.

The shell theorem does not state that the field from a uniformly charged shell is identical to that from the same total charge all located at the center point.