# Two touching neutral conductors in electric field separate?

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1. Aug 9, 2015

### T. Centa

1. The problem statement, all variables and given/known data

Given a situation like this:

Where the two blocks are perfectly conducting materials and are touching, will they separate? If they do separate, will they have the net charge values you calculate in the initial setup using Gauss's Law?

There are no numbers, it's just conceptual.

2. Relevant equations

Gauss' Law

3. The attempt at a solution

I start by treating the two touching blocks as a single rectangular conductor. Then negative charge will accumulate at the top end, and positive at the bottom, and can calculate how much using Gauss's Law. So the two blocks should have a net charge and given that they have a net charge, I figure they should both have a net force on them due to the electric field. I know that there will also be interplay between the accumulated charges (the two ends will attract each other), but for this case assume the heights of the blocks are long enough so that the electric field strength is much stronger than the electric field due to the accumulated charges at each end.

Is this correct, or am I missing something?

2. Aug 9, 2015

3. Aug 9, 2015

### T. Centa

When you put the neutral conductors in the electric field, using Gauss's Law charge will accumulate on the two ends (top end as in the surface on the top conductor closest to the positively charged plate and bottom end as in the surface on the bottom conductor closest to the negatively charged plate). Because charge accumulates on each end, each conductor should have its own net charge (use the picture), unless charge is accumulating elsewhere, but using Gauss's Law, there should be not net charge inside the conductor because there is no electric field. Also, there should be no net charge on the left and right sides as these surfaces are parallel to the electric field. So unless there is something happening at the intersection between the conductors (where they are touching), there should only be net charge at the very top end and very bottom end.

The forces are electric forces between the charged plates and the net charge accumulating on each end which pushes the ends in opposite directions. The top end, which should have a net negative charge, will be pulled upwards in the page while the bottom end, which should be positively charged, will be pulled towards the bottom in the page. Also the electric force between the net charge on the top end (which is negative) will attract the net charge on the bottom end (which is positive) so that will pull the two ends together.

4. Aug 10, 2015

### BvU

I fully agree. Leads to two questions:
1. Are the two half-blocks attracting each other while there is electrical contact ? How and why ?
2. And is there something interesting happening when the two touching surfaces are becoming separated ?

Not clear to me if you now have decided whether the blocks separate due to the pulling apart, or stay together because the pulling together force wins....

Just so you are aware: I'm not the know-all who lures you towards the right answer. Basically I'm just asking questions, and who knows we can sort this out.
May well be we have to make some semi-quantitative considerations as well: the electric field in the gap between the block end and the large capacitor plate, the energy stored in the electric fields etc.

I do assume we're not to worry about gravity, just electric force.

5. Aug 10, 2015

### T. Centa

I fully agree. Leads to two questions:
1. Are the two half-blocks attracting each other while there is electrical contact ? How and why ?
2. And is there something interesting happening when the two touching surfaces are becoming separated ?

Not clear to me if you now have decided whether the blocks separate due to the pulling apart, or stay together because the pulling together force wins....

Just so you are aware: I'm not the know-all who lures you towards the right answer. Basically I'm just asking questions, and who knows we can sort this out.
May well be we have to make some semi-quantitative considerations as well: the electric field in the gap between the block end and the large capacitor plate, the energy stored in the electric fields etc.

I do assume we're not to worry about gravity, just electric force.[/QUOTE]

Ya don't worry about gravity. I still believe that they would be pulled apart, I was just acknowledging that there is an attractive force between them due to the two net charges on each end being oppositely charged. However, if that is the only attractive force between them, then by making the two blocks long enough (ie. separating the two ends a further distance), that attractive force should be overcome by the constant electric field. Keeping in mind that this is a theoretical situation and I can make the blocks as long as I want.

If you would like my thoughts on when they separate: when considering the two blocks touching and assuming they act as a single rectangular conductor would (which I'm hoping is an OK assumption), the electric field in the middle of the two would be zero. Now if you imagine them being pulled an infinitesimal distance apart, given the same charge distribution as when they were touching, the electric field should still be extremely close to zero in that middle space between the two conductors (because the charges barely moved). Given that situation, I don't believe the charge distribution would change much to achieve an electric field of zero within the now separate conductors. If that is indeed the case then the blocks may just continue to separate apart until there is some significant distance between them. But that is just my thought.

I am not sure if there is a quantitative way to determine that. As far as all my searching has come up with, determining the charge distribution when two conductors are present is extremely difficult because the electric fields they produce interact with each other. Unless you know a way to calculate it.

6. Aug 10, 2015

### Qwertywerty

My thoughts - The blocks should remain in contact , or in any case , at most , at an infinitesimally small distance apart .

When the two half - blocks do start to seperate , charges will automatically be induced at the lower end of the upper block , and the upper end of the lower block .

As they both are conducting , the charge on them will now ( finally , after seperation ) rearrange such that charge distribution becomes uniform , and they will effectively act as neutral species on seperation . Thus , now the force due to the charged plates will be zero on both , and they will ultimately have reached a state of equilibrium , some distance apart .

7. Aug 10, 2015

### Qwertywerty

Yeah , I don't get this . Rearrangement takes place with the objective of cancelling electric field inside the the conductor .

And if this does happen , then force should be zero no matter how long the blocks are .

8. Aug 10, 2015

### BvU

Is this a thought exercise or genuine homework ? Cause it's nice to continue this: the blocks hit the plates, discharge and even get charged, so now they attract each other! And the song and dance goes on until the plates are discharged.

9. Aug 10, 2015

### BvU

10. Aug 10, 2015

### T. Centa

What exactly requires the charge distribution to be uniform? Just being conductors does not mean the charge distribution has to be uniform.

11. Aug 10, 2015

### T. Centa

O yeah ? But on the large plates from the original capacitor there is an equal an opposite charge ... so is there really a field pulling on some charge (that we agreed wasn't there -- and you re-state that below) ?

Sorry I am not sure what you mean by this. The field I am talking about is the field produced by the induced charge on the conducting block surface and the other ends induced charge will be in that electric field so will be pulled toward it and vice versa.

So then the blocks hit the plates and charges annihilate each other, right ?

I suppose this would happen if the plates are conducting. Although I assumed they were insulating, that way you can ensure that the charge density remains uniform and the electric field would be constant and perpendicular. It always seems that calculations are just more complicated when its a conductor....

Although yes I would expect the blocks to hit the plates.

Well, a condicting block does something a dielectric ( and here) also does: it reduces E in (a part of the) volume

Can you explain this to me? Because dielectrics, I thought, would not react to an electric field (and therefore cannot reduce it) because their charge distribution will not change.

Is this a thought exercise or genuine homework ? Cause it's nice to continue this: the blocks hit the plates, discharge and even get charged, so now they attract each other! And the song and dance goes on until the plates are discharged.

I am actually trying to build a machine for fun and this is the last piece to the puzzle. So think of it as more of a thought exercise.