# Force between the plates of a charged capacitor

• Jahnavi
In summary, the electric field between the capacitor plates is uniform and confined within the plates. There is no electric field outside the capacitor.
Jahnavi

## 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 .

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

You are right .

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

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

rude man said:
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 ?

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.

Jahnavi said:
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.

rude man said:
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.

Jahnavi said:
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 said:
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.

Orodruin said:
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.

rude man said:
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?

gneill said:
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 .

Jahnavi said:
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.

Jahnavi said:
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?

rude man said:
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 .

Jahnavi said:
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 ...

Jahnavi
Jahnavi said:
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
rude man said:
cognoscenti

Thats a new word I have learned today

gneill said:
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.

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

Would you go for option 2) ?

Jahnavi said:

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!)

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 ?

Jahnavi said:
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

Jahnavi
Jahnavi said:
Thats a new word I have learned today
They're all over here at PF!

Jahnavi
gneill said:
Well, it's at least an argument for a specific case

I support your choice given the poorly formulated problem

Thanks !

gneill said:
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.

gneill said:
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.

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!

cnh1995
rude man said:
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.

## 1. What is the force between the plates of a charged capacitor?

The force between the plates of a charged capacitor is the electrostatic force caused by the attraction or repulsion of the opposite or like charges on the plates. This force is what keeps the plates separated and creates a potential difference between them.

## 2. How is the force between the plates of a charged capacitor calculated?

The force between the plates of a charged capacitor can be calculated using the formula F = Q1Q2/4πεr², where Q1 and Q2 are the charges on the two plates, ε is the permittivity of the medium between the plates, and r is the distance between the plates.

## 3. What factors affect the force between the plates of a charged capacitor?

The force between the plates of a charged capacitor is affected by the magnitude of the charges on the plates, the distance between the plates, and the permittivity of the medium between the plates. It is also affected by the dielectric material between the plates, which can increase or decrease the force depending on its properties.

## 4. How does the force between the plates of a charged capacitor change with charge and distance?

The force between the plates of a charged capacitor follows an inverse square law, meaning that it decreases as the distance between the plates increases. It also increases with the magnitude of the charges on the plates. So, if the distance between the plates is halved, the force will quadruple, and if the charges on the plates are doubled, the force will also double.

## 5. Can the force between the plates of a charged capacitor be repulsive?

Yes, the force between the plates of a charged capacitor can be repulsive if the charges on the plates are of the same sign. In this case, the plates will push away from each other due to the repelling electrostatic force. However, in most cases, the force between the plates is attractive due to the opposite charges on the plates.

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