Force between the plates of a charged capacitor

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Force between capacitor plates is Q²/(2Aε₀) .

If the distance between the plates is small as compared to the area , then force between the plates does not depend on the distance .

But I am not sure how to deal with this problem .

 

[Orodruin]

You are asked for the force between the capacitors, not the force between the plates of a single capacitor.

 

[Jahnavi]

You are right .

And how to deal with that ? Do capacitors behave like dipoles ?

 

[rude man]

Seems to me you need to define how the two capacitors are juxtaposed. Side-by-side? On top of another? Strange question.

 

[Jahnavi]

Side-by-side? On top of another?

How would you deal if it is the former ? How would you deal if it is the latter ? By doing that do you get one of the options ?

 

[gneill]

Can you describe the electric field produced by a charged capacitor? You may want to look through your textbook or class notes to be sure.

 

[rude man]

How would you deal if it is the former ? How would you deal if it is the latter ? By doing that do you get one of the options ?

I suggest that the 4 plates had better be stacked one on top of the other. So you have 4 plates on top of one another as follows: plate, d, plate, L, plate, d, plate.
Determine the 8 surface charge densities on all four plates. By reasoning you can determine that 4 equations and 4 unknowns are adequate to describe the system. Once you know these densities you know the densities of the two surfaces separated by L. From that you know the D (or E) field in the L gap. Use virtual work to get your answer.

 

[Orodruin]

Seems to me you need to define how the two capacitors are juxtaposed. Side-by-side? On top of another? Strange question.

I do not think this is necessary to answer the question. The question only asks for the scaling with d and L, not for a computation of the actual force.

Do capacitors behave like dipoles ?

Is the net charge of the capacitors non-zero? If it is then they behave as monopoles, if it is not then the leading contribution for large L is a dipole contribution.

 

[rude man]

I suggest that the 4 plates had better be stacked one on top of the other. So you have 4 plates on top of one another as follows: plate, d, plate, L, plate, d, plate.
Determine the 8 surface charge densities on all four plates. By reasoning you can determine that 4 equations and 4 unknowns are adequate to describe the system. Once you know these densities you know the densities of the two surfaces separated by L. From that you know the D (or E) field in the L gap. Use virtual work to get your answer.

Unfortunately I come up with the E field = 0 in the L gap, as is the attractive force. The only attraction forces I see are those between each plate pair, with L being immaterial.

 

[rude man]

Is the net charge of the capacitors non-zero? If it is then they behave as monopoles, if it is not then the leading contribution for large L is a dipole contribution.

Each capacitor's plate pair must have equal and opposite charge; that's the only way to "charge a capacitor". So neither capacitor can have net charge.

 

[Orodruin]

Each capacitor's plate pair must have equal and opposite charge; that's the only way to "charge a capacitor". So neither capacitor can have net charge.

Are you the OP? If not why reply to a rhetorical question intended for the OP?

 

[Jahnavi]

Can you describe the electric field produced by a charged capacitor?

Electric field between the plates is uniform ( some fringing occurs near the ends) . It is confined within the plates . There is no electric field outside the capacitor .

 

[rude man]

Electric field between the plates is uniform ( some fringing occurs near the ends) . It is confined within the plates . There is no electric field outside the capacitor .

Just wanted to add that my conclusions were based on equal and opposite charge on each capacitor plate pair since that's how capacitors are usually "charged" (a misnomer anyway) i.e. by running current thru them.

 

[rude man]

There is no electric field outside the capacitor .

OK if there is no E field between the capacitors, how can there be a force between them? Or am I misinterpreting the problem totally?

 

[Jahnavi]

OK if there is no E field between the capacitors, how can there be a force between them? Or am I misinterpreting the problem totally?

I don't know

@gneill , I have replied to your question . Please see .

 

[rude man]

I don't know
@gneill , I have replied to your question . Please see .

Hard question! I think we can rule out magnetism, gravity, and the weak and strong interactions ...
Perhaps one of our cognoscenti will report in ...

 

[gneill]

I don't know

@gneill , I have replied to your question . Please see .

My intuition would lead me to believe that there would be no force acting on the pair of the relatively widely separated capacitors, since theory tells us that there is essentially no E-field external to the space between the plates. However, the problem's selection of answers does not include that choice. This leads me to think that the problem's author was thinking that the charges on the plates are to be viewed (from a distance) as point charges forming dipoles, although I question that interpretation.

If that is the case, then you should look into the interaction of dipoles. A web search will turn up relevant material that will let you choose between the given answers, at least for suitably oriented and positioned dipoles. In my opinion the problem as stated is altogether too vague in its description of the scenario.

 

[Jahnavi]

cognoscenti

Thats a new word I have learned today

 

[Jahnavi]

My intuition would lead me to believe that there would be no force acting on the pair of the relatively widely separated capacitors, since theory tells us that there is essentially no E-field external to the space between the plates. However, the problem's selection of answers does not include that choice. This leads me to think that the problem's author was thinking that the charges on the plates are to be viewed (from a distance) as point charges forming dipoles, although I question that interpretation.

If that is the case, then you should look into the interaction of dipoles. A web search will turn up relevant material that will let you choose between the given answers, at least for suitably oriented and positioned dipoles. In my opinion the problem as stated is altogether too vague in its description of the scenario.

Thanks for replying .

I did mention about capacitors behaving like dipoles in post#3 .

Would you go for option 2) ?

 

[gneill]

Thanks for replying .

I did mention about capacitors behaving like dipoles in post#3 .

Would you go for option 2) ?

I might if you can justify that choice with a suitable argument
(We can't confirm or deny what might be guesses!)

 

[Jahnavi]

The force between two dipoles when the axes are aligned along the same straight line is given by 6kp1p2/r⁴ .

This gives option 2) .

Is that a suitable argument ?

 

[gneill]

The force between two dipoles when the axes are aligned along the same straight line is given by 6kp1p2/r⁴ .

This gives option 2) .

Is that a suitable argument ?

Well, it's at least an argument for a specific case

I support your choice given the poorly formulated problem

 

[rude man]

Thats a new word I have learned today

They're all over here at PF!

 

[Jahnavi]

Well, it's at least an argument for a specific case

I support your choice given the poorly formulated problem

Thanks !

 

[Orodruin]

This leads me to think that the problem's author was thinking that the charges on the plates are to be viewed (from a distance) as point charges forming dipoles, although I question that interpretation.

Why do you question this? The monopole contribution for an overall neutral system is zero and so the leading contribution must be the dipole-dipole interaction.

 

[rude man]

Well, it's at least an argument for a specific case

I support your choice given the poorly formulated problem

So do I. And it can be shown that indeed the orientation of the dipoles does not affect the answer. Thanks to you & orodruin for pointing out the idea of dipoles.

 

[rude man]

Last shot: usually we assume that parallel-plate capacitors have no fringing field. In this case the opposite is true: the fringing field is the whole story!

 

[Orodruin]

Last shot: usually we assume that parallel-plate capacitors have no fringing field. In this case the opposite is true: the fringing field is the whole story!

Indeed. That is because it falls of as 1/r^3 due to the dominant pole being the dipole rather than 1/r^2 so it becomes negligible much faster than your typical monopole field.