Energy ratio of two systems with charged planes

[Eitan Levy]

Homework Statement

In those systems, the charge of the exterior planes is Q and the charge of the interior plane is is -2Q. Ignore side effects.
Find the ratio E(B)/E(A), where B is the system from the right.

Relevant Equations

E=σ/2ε[SUB]0[/SUB]

Honestly no idea how to get an answer. I found the electric field between the planes and out of it in those two cases but this didn't take me very far.

The final answer is E(B)/E(A)=2. Can someone please explain why?

 

[gneill]

Can you show some details for what you've done so far? What field values have you found for the exteriors (and how did you determine them)?

 

[TSny]

Please give the complete statement of the problem. Are the plates conductors?

It looks like you might be using the symbol E for both energy and electric field.

 

[gneill]

Please give the complete statement of the problem. Are the plates conductors?

It looks like you might be using the symbol E for both energy and electric field.

@TSny : Ha! Good question. I missed that. The thread title mentions energy and I immediately forgot that when I read the problem statement. I assumed that the E's represented the electric fields.

The two setups represent capacitor configurations, and one could try to evaluate their stored energy. The case on the left is relatively straightforward, but I'm not sure that "Ignore side effects" really covers all sins here when looking at the right-side case. Your comment about the plates being conducting is telling.

 

[TSny]

@TSny :
The two setups represent capacitor configurations, and one could try to evaluate their stored energy. The case on the left is relatively straightforward, but I'm not sure that "Ignore side effects" really covers all sins here when looking at the right-side case. Your comment about the plates being conducting is telling.

Yes, if the plates are conductors, then you can make certain idealized approximations which lead to the stated answer for the ratio of the electrostatic energy in the two systems. I haven't worked it based on modeling the systems as capacitors. But, that should be a nice way. I was thinking of the energy as stored in the electric fields between the plates.

 

[gneill]

I think we need to hear back from the OP to ascertain the meaning of "E" in the question.

 

[Eitan Levy]

I think we need to hear back from the OP to ascertain the meaning of "E" in the question.

My apologies, the question is discussing the ratio of energies.

 

[rude man]

My apologies, the question is discussing the ratio of energies.

Conductors or dielectrics?

 

[Eitan Levy]

Conductors or dielectrics?

Conductors. I know it has been a while, but I still don't know how to get an answer to this problem.

 

[rude man]

1. I assume you know how to determine the E fields in system A.

System B:
realize you can split this into two independent subsystems. You can cut each plate into 2 equal plates of area LW/2 and physically separate the top 2 half-plates from the 3 lower half-plates. So you have two half-plates separated by 2d, and 3 half-plates each separated by d; each system of half-plates is lined up symmetrically. (A half-plate has area = LW/2). The two top plates have nothing to do with the bottom 3 except for the fact that the total charge on each full-plate is as given.

Now treat the two systems as totally isolated from each other and compute the various $\sigma$ (surface charge densities) for each system.

Once you have all the sigmas you know the top & bottom electric fields $E$, then you can compute the two energies by $U = 1/2~~D \cdot E, D = \epsilon_0 E$..

 

[Eitan Levy]

1. I assume you know how to determine the E fields in system A.

System B:
realize you can split this into two independent subsystems. You can cut each plate into 2 equal plates of area LW/2 and physically separate the top 2 half-plates from the 3 lower half-plates. So you have two half-plates separated by 2d, and 3 half-plates each separated by d; each system of half-plates is lined up symmetrically. (A half-plate has area = LW/2). The two top plates have nothing to do with the bottom 3 except for the fact that the total charge on each full-plate is as given.

Now treat the two systems as totally isolated from each other and compute the various $\sigma$ (surface charge densities) for each system.

Once you have all the sigmas you know the top & bottom electric fields $E$, then you can compute the two energies by $U = 1/2~~D \cdot E, D = \epsilon_0 E$..

I am unable to calculate the densities.

I tried to look at the system like this:

I really don't know if it's a correct analysis of the system, and even if it is I don't know how to proceed. Are the charges on the second and third plates (looking at the lower sub-system) -2q? Are they -2q combined? Can I assume the that each pair has opposite densities? I really don't know.

 

[rude man]

I don't think looking at capacitor equivalents is a good way to go. Just my opinion. Capacitors usually have equal & opposite charge on their plates since they're charged or discharged by a current. But some folks make it work sometimes.

I prefer to assign $\sigma$ to each surface. So for the top plates you have, left to right,$\sigma_1, _2, _3 ~and~_ 4.$ You can eliminate some of these; for example, what about $\sigma_2$ and $\sigma _3$? And you know that $\sigma_1+\sigma_2$ must equal the charge on that half-plate. You also know that for a given plate, charge on top half-plate + charge on bottom half-plate must equal the given total charge. And so on. You need to play around with these concepts, get a system of equations to solve for all the $\sigma$.

 

[Eitan Levy]

I don't think looking at capacitor equivalents is a good way to go. Just my opinion. Capacitors usually have equal & opposite charge on their plates since they're charged or discharged by a current. But some folks make it work sometimes.

I prefer to assign $\sigma$ to each surface. So for the top plates you have, left to right,$\sigma_1, _2, _3 ~and~_ 4.$ You can eliminate some of these; for example, what about $\sigma_2$ and $\sigma _3$? And you know that $\sigma_1+\sigma_2$ must equal the charge on that half-plate. You also know that for a given plate, charge on top half-plate + charge on bottom half-plate must equal the given total charge. And so on. You need to play around with these concepts, get a system of equations to solve for all the $\sigma$.

This is where I am getting confused at:

The electric field inside the conductors must be equal to 0. So if I take a Gauss pillbox between an outer plate and the middle plate (looking at the lower part), the densities there must cancel each other. That means that on the lower parts of the outer plates there is a charge of 2q. That leaves me with -q for the upper parts but it doesn't make sense because then taking the same pillbox the electric field inside the conductors will not be 0.

Edit: I'm trying to do what Kuruman did here in comment number 2: https://www.physicsforums.com/threa...ity-of-an-infinite-plate.994711/#post-6403647

 

[rude man]

This is where I am getting confused at:

The electric field inside the conductors must be equal to 0. So if I take a Gauss pillbox between an outer plate and the middle plate (looking at the lower part), the densities there must cancel each other. That means that on the lower parts of the outer plates there is a charge of 2q.

Why? The middle plate has a total charge of 2Q so half that charge, i.e. Q, could reside on each surface, in fact on the upper half of each surface if the lower half has zero charge; and so you'd get equal & opposite charge on each lower outer half-plate

 

[Eitan Levy]

This is where I am getting confused at:

The electric field inside the conductors must be equal to 0. So if I take a Gauss pillbox between an outer plate and the middle plate (looking at the lower part), the densities there must cancel each other. That means that on the lower parts of the outer plates there is a charge of 2q. That leaves me with -q for the upper parts but it doesn't make sense because then taking the same pillbox the electric field inside the conductors will not be 0.

Do you mean that on one side of the middle plate there would be charge -q and on the other -q again? The concept of sides when talking about planes is very confusing to me. How did you know the charge would split evenly between the two "sides"?

If you find the time please look at the other thread in which you helped, I explained my problem further there.

 

[rude man]

Do you mean that on one side of the middle plate there would be charge -q and on the other -q again? The concept of sides when talking about planes is very confusing to me. How did you know the charge would split evenly between the two "sides"?

It should be obvious from symmetry considerations. How could it be otherwise?

If you find the time please look at the other thread in which you helped, I explained my problem further there.

I will, but I think I'm offering you a good, systematic way of solving this kind of problem once you learn it. You would be able to solve any arbitrary charge distribution of any number of plates, not only symmetrical ones. There are only 2 steps, and they don't involve gaussian 'pillboxes'. Or capacitors.

 

[Eitan Levy]

Why? The middle plate has a total charge of 2Q so half that charge, i.e. Q, could reside on each surface, in fact on the upper half of each surface if the lower half has zero charge; and so you'd get equal & opposite charge on each lower outer half-plate

Well, then I assume that for the outer plates the charge is entirely on the lower part. The only way I am able to understand this is by thinking about the upper part of the middle plates and the lower part of each outer plate to be capacitors. So if each side of the middle plate has the charge -q, the other parts of the capacitors (i.e. the lower parts of the outer plates) must have a charge q.

Please, how can you deduce otherwise that the charge on the lower outer plates is q? I would love to hear.

 

[rude man]

Well, then I assume that for the outer plates the charge is entirely on the lower part.

Right.
You should be aware that for an electrostatic field, flux lines begin at one polarity of charge and end at the other. By symmetry, charge on the upper half of the outer plates must have the same polarity. Which says what about the E field between the upper outer plates? E.g. using your pillboxes?

The only way I am able to understand this is by thinking about the upper part of the middle plates and the lower part of each outer plate to be capacitors. So if each side of the middle plate has the charge -q, the other parts of the capacitors (i.e. the lower parts of the outer plates) must have a charge q.

That reasoning is correct but it is unnecessary to think capacitors.
What you learned about capacitors is based on basic electrostatics, not some gizmo for which i = C dV/dt.

Please, how can you deduce otherwise that the charge on the lower outer plates is q? I would love to hear.

2 things:
1. $Q = \Sigma \sigma_i~A_i$ where
$\sigma$ is surface charge density,
$A = area$
$Q =$ total charge on a plate.

2. The zero E field in a conductor is ascribable to what? Think a test charge inside the conductor. What forces sum up to zero if there are charges near that conductor?

Using those 2 facts you can handle any coinfiguration of plates, any charge distribution.

Gedankenexperiment for you: Put a piece of metal in the gap of a paralle-plate charged capacitor, not touching either plate. Why is the E field in the metal zero? What happened to the external capacitor E field $E = \sigma/\epsilon$ ??

 

[Eitan Levy]

Right.
You should be aware that for an electrostatic field, flux lines begin at one polarity of charge and end at the other. By symmetry, charge on the upper half of the outer plates must have the same polarity. Which says what about the E field between the upper outer plates? E.g. using your pillboxes? That reasoning is correct but it is unnecessary to think capacitors.
What you learned about capacitors is based on basic electrostatics, not some gizmo for which i = C dV/dt.
2 things:
1. $Q = \Sigma \sigma_i~A_i$ where
$\sigma$ is surface charge density,
$A = area$
$Q =$ total charge on a plate.

2. The zero E field in a conductor is ascribable to what? Think a test charge inside the conductor. What forces sum up to zero if there are charges near that conductor?

Using those 2 facts you can handle any coinfiguration of plates, any charge distribution.

Gedankenexperiment for you: Put a piece of metal in the gap of a paralle-plate charged capacitor, not touching either plate. Why is the E field in the metal zero? What happened to the external capacitor E field $E = \sigma/\epsilon$ ??

I am sorry but I don't understand. I really don't know at all how to analyze this.

If you are willing to do so please write how to analyze this probably. It will help me tremendously, I have already wasted hours on trying to understand this.

EDIT: Let me elaborate.

Let me elaborate.

If the charges are equal on the top parts then the total field there is 0. No energy is stored there.

Usually the field inside conductors is 0 because there is movement of electrons twhich cancel out the outer field.

Still, I don't know how to use these facts to analyze the system.