The electric field in an expanding, statically-charged tube

In summary: I'm just a test charge.In summary, the tube appears to expand slowly perpendicular to the x-axis while remaining centered on the x-axis. The charge experiences a magnetic field and an electric field, and the electric field depends on the velocity of the charge relative to the tube.
  • #1
particlezoo
113
4
Imagine that I have a straight, statically-charged, cylinder-shaped tube with arbitrary (ideally infinite) extent. The charge is distributed evenly over the tube such that the field inside the tube is zero. For convenience, let's line up the tube centered along the x-axis such that the "centerline" of the tube is exactly on the x-axis. Ok, so let's say I permit this tube to expand or contract slowly perpendicular to the x-axis, while remaining centered on the x-axis. The tube is at rest in the inertial frame where the coordinate system is fixed.

The following questions pertain to what is observed in this rest frame:

Question 1) Is it correct that there will be no electric field at the x-axis?
Question 2) Is it correct that there will be no magnetic field at the x-axis?

If the answer to both questions is yes, then let's continue.

-------------------------------

Let's say I'm a test charge. My charge is the same sign as the charge of the tube. I am on the x-axis, and I am moving in the positive x direction. There is no y or z component to my movement. I have a very steady velocity and will not accelerate substantially because I am extremely massive.

What do I see?

I think I see the charged tube passing by me at some velocity. While I do not experience any net magnetic field, I can infer that there is a magnetic field outside the tube. I can also infer that my electric field, though quite small in strength as I am a simple test charge, nevertheless adds to the electric field outside the tube and therefore there must exist some additional energy stored in the electric field due to my field adding to the electric field outside the tube.

The tube appears to expand ever so slowly, but I do see that it is expanding. The walls are moving away from me. What do I see when the tube is expanding?

I think I see that the magnetic field outside the tube is mostly unchanged, except that the radius of the tube is changing, so the nearest magnetic field which was in a space closer to the x-axis (that is now inside the tube) no longer exists. When I consider that my electric field crossed with this magnetic field carried some amount of field momentum in the form of E x B (where E stands for the electric field and B stands for the magnetic field), it appears that the amount of momentum in the field has changed. It would follow that something must be getting the momentum that this field lost. Hmm... I have a question:

Question 3) From my perspective as the little test charge, do I see an electric field which attempts to reduce my relative velocity to this expanding tube?

Again, I am just a little test charge, but I'd like to imagine what it would be like to be in this tube's frame of reference. What does it look like? In the tube's rest frame, it doesn't seem that this tube would produce any magnetic field, but it appears that I would generate a magnetic field of my own. This magnetic field that I have would interact with the electric field of the tube, and as the tube expands, the momentum of this field would change. It seems that the magnetic field that I possesses is pointed in the direction opposite of what I saw versus when I was myself moving through the tube, where I saw (or inferred rather) the tube's magnetic field. Since the sign of my charge is the same as that of the tube's, it appears that this interaction field momentum does not point the same way that I saw in my rest frame, but in the opposite direction. I have another question:

Question 4) When I imagine looking at myself from outside the tube, in the tube's rest frame, do I see myself subject to an electric field which attempts to reduce my relative velocity to this expanding tube?

I wonder...

-------------------------------

Ok, so what do think about that guys? My major concern is that, in the rest frame of the tube, the electric field from the expanding tube's charged walls does not exist at the x-axis.

Also, the electric field that the (extremely massive) test charge experiences in its rest frame appears to depend on the relative x-velocity it has with the expanding tube. In an alter-scenario with the reverse (and still very steady) relative x-velocity, the force on the test charge in its (alter-scenario) rest frame would be the reverse.

So in the tube's rest frame what exerts the force on the charge as should be expected from Lorentz transformations? Is it a field? In the tube's rest frame, it doesn't seem there's any field that the tube produces inside itself, let alone one that depends on the velocity of the charge. Scientific literature says that only fields cause force on charges and that potentials not really responsible (i.e. potentials are not needed to explain forces on the charges). That basically rejects the idea that a time-varying potential acts on a moving particle to provide the missing force. Anyway, we are dealing with classical physics here, so I am looking for a valid resolution in the realm of classical physics only.
Kevin M.
 
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  • #2
You have too many conditional questions and too many possible misinterpretations.

I believe it will make a difference if the tube is conductive, or if the charges on the tube are locked in place on the surface of an insulated tube. Which is it?

What is your reference? Outside the tube it appears to be the environment at infinity. Inside the tube, is it the wall of the tube or the axial filament of the test charge. If the wall is conductive then the inside and outside environments cannot be connected, they are quite separate spaces or environments. It is then not possible for one side to sense the other universe.

By changing the tube diameter you are changing the capacitance per unit length to the external environment. The external charge on the tube is fixed, so the tube voltage must change, but that is relative to infinity. C = Q / V.

As a test charge on the x-axial current filament, you may see your own inverted image in the conductive wall mirror. The axial line of test charge would see a tube of twice the radius and an opposite charge. Changing the tube diameter will change the diameter of the image and the internal coaxial capacitance per unit length. That will change the test charge to wall difference voltage.

The test charge may then behave like a signal inside a tapered coaxial transmission line. The impedance will change as the test charge progresses, so charge will be progressively reflected while the voltage rises and the current falls.
 
  • #3
Baluncore said:
You have too many conditional questions and too many possible misinterpretations.

I believe it will make a difference if the tube is conductive, or if the charges on the tube are locked in place on the surface of an insulated tube. Which is it?

The scenario I have in mind has an insulated cylindrical tube where the charge density is fixed and uniform on the surface so as to minimize the internal electric field in the tube's rest frame. An arbitrarily-long or infinite-length tube would contribute zero electric field inside in the tube's rest frame, so my thinking goes. Only the tube's rest frame and the (charge on the x-axis)'s rest frame are considered. In the tube's rest frame, the observer is placed in any arbitrary point inside or outside the tube, while in the rest frame of the (charge on the x-axis), the observer is placed at the position of the (charge on the x-axis).

The main issue is essentially the question of whether the charge on the x-axis should see (i.e. whether an observer in the charge's rest frame at the position of the charge should see) an electric field acting on it.

In contrast, I expect that in the tube's rest frame, the only electric field inside the tube (if infinitely-long with evenly distributed surface charge) will be from the charge on the x-axis.
 
  • #4
particlezoo said:
The scenario I have in mind has an insulated cylindrical tube where the charge density is fixed and uniform on the surface so as to minimize the internal electric field in the tube's rest frame. An arbitrarily-long or infinite-length tube would contribute zero electric field inside in the tube's rest frame, so my thinking goes. Only the tube's rest frame and the (charge on the x-axis)'s rest frame are considered. In the tube's rest frame, the observer is placed in any arbitrary point inside or outside the tube, while in the rest frame of the (charge on the x-axis), the observer is placed at the position of the (charge on the x-axis).

The main issue is essentially the question of whether the charge on the x-axis should see (i.e. whether an observer in the charge's rest frame at the position of the charge should see) an electric field acting on it.

In contrast, I expect that in the tube's rest frame, the only electric field inside the tube (if infinitely-long with evenly distributed surface charge) will be from the charge on the x-axis.

The reason why I expect that a field would be seen by the charge on the x-axis is based on what a charge would see from a wire approaching or receding from it:

perpend12.gif

Electric field from current in wire as it approaches a charge.

Similar but different to the above picture is our scenario in question where we consider what the charge on the x-axis "sees". In this case, there is excess charge on the tube enveloping the x-axis that is fixed in place, so there is no embedded current to speak of (i.e. no relative motion of positive and negative charges on the insulated tube). However, there would be excess charge of one type vs. another type, so a diagonal contribution would be expected from each "parallel strand" of the tube as the tube changes radius. The electric field contributions at the charge from all the "parallel strands" together would reduce to a "longitudinal" component on the x-axis, with all the radial(perpendicular) components cancelling at the x-axis. If my suspicions are correct, this electric field observed would depend on the relative velocity between the tube and the charge.
 

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  • #5
particlezoo said:
The scenario I have in mind has an insulated cylindrical tube where the charge density is fixed and uniform on the surface so as to minimize the internal electric field in the tube's rest frame. An arbitrarily-long or infinite-length tube would contribute zero electric field inside in the tube's rest frame, so my thinking goes.
So the tube is a dielectric material. The introduction of a test charge will cause a displacement current to move some charge from one side of the tube wall towards the other surface.
 
  • #6
Baluncore said:
So the tube is a dielectric material. The introduction of a test charge will cause a displacement current to move some charge from one side of the tube wall towards the other surface.

Wouldn't that be dependent on the electric susceptibility of the material, which could be close to zero in some variation of this scenario, thereby closely approximating a tube with charge "fixed" to its walls?
 
  • #7
Susceptibility, Xe + 1 = Er relative permittivity, which cannot be less than free space = 1.
If charges are present then the relative permittivity must be greater than one.

Zero susceptibility is free space. A vacuum capacitor, filled with free space, still has capacitance and a displacement current at the surface of the plates.

You cannot deny capacitance when charges are held by a fixed insulator, it is a dielectric.
If the insulator was so good that it had Xe = 0 then it would be free space and the charges would not be held, they would migrate.

The idea that there can be no charge on the inside of a tube only goes for closed conductive tubes.
 
  • #8
Baluncore said:
Susceptibility, Xe + 1 = Er relative permittivity, which cannot be less than free space = 1.
If charges are present then the relative permittivity must be greater than one.

Zero susceptibility is free space. A vacuum capacitor, filled with free space, still has capacitance and a displacement current at the surface of the plates.

You cannot deny capacitance when charges are held by a fixed insulator, it is a dielectric.
If the insulator was so good that it had Xe = 0 then it would be free space and the charges would not be held, they would migrate.

The idea that there can be no charge on the inside of a tube only goes for closed conductive tubes.

I agree, but I am not sure this fully addresses my concern about the potential discrepancy of what the charge on the x-axis sees in its rest frame vs what we see in the tube's frame. First, the polarization current, or the induced current in the dielectric, is not strictly proportional to the velocity of the test charge and its magnitude of its charge. So if you say that in the rest frame of the tube a polarization current is induced in the dielectric due to the relative motion of the charge on the x-axis, the induced electric field on the x-axis not only depends on the charge magnitude and its velocity, but also the susceptibility of the dielectric. We should see how this compares to the electric field as seen in the rest frame of the charge.

As far as what the charge on the x-axis sees in its rest frame, it depends not only on the relative velocity and the magnitude of the charge of the tube (and its distribution which is modified by the polarization current), but also the expansion of the tube. So for a tiny test charge compared to say a tube with a very large charge density on its surface, it seems that the polarization current would lead to quite small changes (percentage wise) in the resultant electric field that the test charge would experience. This would be the case even if the electric susceptibility were quite high.
 
  • #9
particlezoo said:
So if you say that in the rest frame of the tube a polarization current is induced in the dielectric due to the relative motion of the charge on the x-axis, the induced electric field on the x-axis not only depends on the charge magnitude and its velocity, but also the susceptibility of the dielectric.
NOT the relative motion nor the velocity. The movement of charges in the tube wall is due to the presence of the axial charge. A lump of displaced charge will surround a moving axial test charge. That lump will be proportional to and will follow the test charge along the cylindrical capacitor. Remember the electric field of the test charge propagates at the speed of light.
The susceptibility of the dielectric tube changes the relative dielectric constant. That changes the capacitance and therefore the voltage of the cylindrical capacitor. You can ignore the tube material susceptibility as it is a constant for your imaginary tube, a tube that has zero thickness is not there. You appear to be making a post modernist terminology soup.
 
  • #10
Baluncore said:
NOT the relative motion nor the velocity. The movement of charges in the tube wall is due to the presence of the axial charge.

Presence reads to me a "placement", implying that the charge is only "placed" and that therefore a "location" is given, but not imply a velocity. The movement of charges that results of solely of "presence" implies then only a transient movement of charges, not a sustained current. Without a velocity, I could not say which direction the charges are displaced more than the other, as they could be displaced just as easily left or right, and then I would not be able to say which direction the electric field from that charge distribution should point along the axis.

Baluncore said:
A lump of displaced charge will surround a moving axial test charge. That lump will be proportional to and will follow the test charge along the cylindrical capacitor. Remember the electric field of the test charge propagates at the speed of light.

I agree completely with this. So there is a displacement current that results in the redistribution of charge as the charge travels. As it does so, it should generate an electric field where the axial charge is at. This happens due to the relative velocity of the charge and the dielectric tube. While the field reaches at the speed of light, the lump travels along the dielectric based on the relative velocity. No relative velocity, then there is no ongoing redistribution to be had.

Baluncore said:
The susceptibility of the dielectric tube changes the relative dielectric constant. That changes the capacitance and therefore the voltage of the cylindrical capacitor. You can ignore the tube material susceptibility as it is a constant for your imaginary tube, a tube that has zero thickness is not there. You appear to be making a post modernist terminology soup.

I never said that it had zero thickness. The tube has a finite radius, and the thickness of the walls were not assumed.

I don't know why, but you seem to think that the redistribution of the charge on the tube is necessary for the charge on the x-axis to see any electric field at all. Since you brought it up though, I can see that you disagreed with me (successfully) about the interior field in the rest frame of the tube caused by charges on the tube, which you pointed out must be distributed non-uniformly. I assumed that there should be no such field, thinking that I could avoid worrying about charge redistribution by implementing a test charge with negligible magnitude (Note: I was under the impression that the term "test charge" should clearly indicate a negligible charge which has a trivial effect on the environment, similar to the idea of a "test mass" when working on a gravitational problem). You pointed out that I must consider charge redistribution in the problem. Okay, I get that. Now, if it is possible, I want to isolate the effects of:

1) The electric field due to an imagined uniform charge distribution on the tube.
2) The electric field due to the induced currents / displaced charge density.

Since fields obey superposition, then it should be possible to say something about what each contributes to the result. I was contriving the problem essentially to justify focusing on 1 and to limit or trivialize the effect of 2. I am under the impression that I can form the problem so I can concern myself with only what the imagined uniform charge distribution would do in the axial charge's rest frame without having to complicate it by considering the effects of charge redistribution, which I am trying to minimize. It seems that you are telling me that I cannot do that in the problem.
 
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  • #11
In case if you missed this post, here it is again from above. It highlights my main concern. Of course here I am only considering what the bulk of the charge on the tube would do (i.e. the uniform contribution) under the guiding assumption that redistribution of charge can be made negligible by making the test charge on the x-axis of negligible value.

particlezoo said:
The reason why I expect that a field would be seen by the charge on the x-axis is based on what a charge would see from a wire approaching or receding from it:

View attachment 213462
Electric field from current in wire as it approaches a charge.

Similar but different to the above picture is our scenario in question where we consider what the charge on the x-axis "sees". In this case, there is excess charge on the tube enveloping the x-axis that is fixed in place, so there is no embedded current to speak of (i.e. no relative motion of positive and negative charges on the insulated tube). However, there would be excess charge of one type vs. another type, so a diagonal contribution would be expected from each "parallel strand" of the tube as the tube changes radius. The electric field contributions at the charge from all the "parallel strands" together would reduce to a "longitudinal" component on the x-axis, with all the radial(perpendicular) components cancelling at the x-axis. If my suspicions are correct, this electric field observed would depend on the relative velocity between the tube and the charge.
 

Related to The electric field in an expanding, statically-charged tube

1. What is an electric field?

An electric field is a physical field that surrounds charged particles and exerts a force on other charged particles within its range. It is represented by electric field lines that indicate both the direction and strength of the force.

2. How does an electric field behave in an expanding tube?

In an expanding, statically-charged tube, the electric field will also expand and increase in strength. As the distance between the charged particles increases, the electric field lines will become more spread out and the force exerted on other charged particles will decrease.

3. What happens to the electric field if the charge in the tube is increased?

If the charge in the tube is increased, the electric field will also increase in strength. This is because there are now more charged particles present to exert a force on other particles within the field.

4. How does the electric field in an expanding tube compare to a stationary tube?

In a stationary tube, the electric field remains constant and does not change with distance. In an expanding tube, the electric field will decrease in strength as the distance between charged particles increases due to the expansion of the tube.

5. Can an expanding, statically-charged tube have a negative electric field?

No, an expanding, statically-charged tube can only have a positive electric field. This is because the electric field is always directed away from the positively charged particles and towards the negatively charged particles. As the tube expands, the direction of the field lines will not change, thus maintaining a positive electric field.

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