Parallel Plate Capacitor Filled with Many dielectrics

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  • #1
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Homework Statement:

Help!

Relevant Equations:

E=(1/Er)*Evac
Evac=(sigma/AEo)(Z component vector)
Screen Shot 2020-05-20 at 7.04.44 PM.png

This is the problem I'm working on. So far I know:
1. I am assuming the free charge density is +sigma for the top plate and -sigma for the bottom plate.
2. The electric field from the plates goes from top to bottom plate, in the negative z direction.
3. The electric field of the capacitors oppose the electric fields of the plates (in +z direction) and are E=1/(Er)*Evac
4. The electric field of the vacuum is Evac=(sigma/AEo)(Z component vector)

I'm unsure about how geometry plays into these questions. From the formula for the electric field in each dielectric, it looks like only the dielectric constant contributes to the electric field for each slab. (since A is always the area of the capacitor plates. Or does that change based on the area that the slab covers?)

I have more questions but I'll just start off with that.
 
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Answers and Replies

  • #2
Paul Colby
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In your class do they talk about the electric field, ##E##, and the electric displacement field, ##D##? These are related by a constitutive relation,

##D = \epsilon E##

Any of this ring any bells?
 
  • #3
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Yes I'm taking Electrodynamics right now over the summer as an independent study and I am covering those topics. Right now, I'm less familiar with D.
 
  • #4
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I'm thinking I should start off by finding the electric field between the plates as if the dielectrics weren't there.
 
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  • #5
Paul Colby
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Okay, capacitance is the charge stored per volt applied. So one needs that charge on one plate for a given voltage.

1) How would you calculate the voltage between a point a on the top plate and a point b on the bottom given the E field? Does this depend on which points are chose on the top and bottom?

2) How would you calculate the charge on the top plate given the D field?

3) What happens to the D field as it passes through the step between dielectrics?
 
  • #6
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Alright so
1) You can find voltage given the electric field by taking the negative of the integral of the electric field from point a to point b. The field is in the z direction, so the voltage shouldn't change if you stay on the plates.
2) If you have the D field, you can take the closed integral over the surface to find the enclosed charge.
3) D field will have a new constant?
 
  • #7
Paul Colby
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Okay, I think the big thing you're missing is that between dielectrics there are no surface charges[1]. This means that as one crosses the boundary between ##\epsilon_1## and ##\epsilon_2##,

##D_1=\epsilon_1E_1 = \epsilon_2 E_2 = D_2##

at least as far as the component of the fields normal to the interface between dielectrics. This means the E-field changes value as you step across.

[1] Okay, there could be but in practice they usually have no way of getting there.

Another thing I find helpful is to remember that E has units of volts/meter where D has units of coulombs/meter^2. D is like a surface charge density.
 
  • #8
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So I calculated my Er1 and got (sigma)/4Eo(Zvector). If I would replace sigma with q/A, then would I use (1/2)A for A, since Er1 only covers half the area of the capacitor?
 
  • #9
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The electric field in Er1 I mean.
 
  • #10
Paul Colby
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There is an E/D field originating at each point on the top plate and ending on the bottom. You already know whatever happens along a single field line that the E's each must integrate to V, the potential drop. Each line travels through some combination of ##\epsilon_i## by the statement of the problem. For each line you must calculate the E's using that the over all D, whatever that ends up being, is constant along a given line. Then sum the charge for each distinct region. There are some equations that you need to setup and solve.
 
  • #11
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Are you talking about for solving capacitance now?
 
  • #12
Paul Colby
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Sure. You assume a potential (applied voltage), V. Solve for E in the various regions. Compute the D value for each region. Sum the charge on the top plate implied by the resulting D for each region of top plate. Divided this total charge by V and you have the capacitance. Problem solved.
 
  • #13
Delta2
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This problem is giving me a headache, what is the fundamental assumption here:
1) D is the same throughout the capacitor interior?
2) E is the same throughout the capacitor interior?
3) None of the above?
 
  • #14
Paul Colby
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This problem is giving me a headache
The original post has a diagram. Half the plate area, ##A##, is a single dielectric, ##\epsilon_3##. The other half a stack of two dielectrics with relative thicknesses, ##L=L_1+L_2##, given. D for a given field line (assume they are all straight and perpendicular to the plates) depends on the dielectric stack it passes through.

So, the system of equations are.

## V = E_1L_1 + E_2L_2##
## D_1 = \epsilon_1 E_1 = \epsilon_2 E_2##
## Q = D_1 A/2 + D_2 A/2##
## D_2 = \epsilon_3 E_3##
## V = E_3 (L_1 + L_2)##

solve these then,

##C = \frac{Q}{V}##

The real problem is understanding the reason for each equation.
 
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  • #15
Delta2
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The original post has a diagram. Half the plate area, ##A##, is a single dielectric, ##\epsilon_3##. The other half a stack of two dielectrics with relative thicknesses, ##L=L_1+L_2##, given. D for a given field line (assume they are all straight and perpendicular to the plates) depends on the dielectric stack it passes through.

So, the system of equations are.

## V = E_1L_1 + E_2L_2##
## D_1 = \epsilon_1 E_1 = \epsilon_2 E_2##
## Q = D_1 A/2 + D_2 A/2##
## D_2 = \epsilon_3 E_3##
## V = E_3 (L_1 + L_2)##

solve these then,

##C = \frac{Q}{V}##

The real problem is understanding the reason for each equation.
I am understanding most of the equations except why as you say "D for a given field line depends on the dielectric stack it passes through" so we split D in D_1 and D_2. The way I understand this problem is that D is one and only.
 
  • #16
Paul Colby
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I am understanding most of the equations except why as you say "D for a given field line depends on the dielectric stack it passes through"
Yeah, no. There are two D's because there are two stacks. The first is ##\epsilon_1## of thickness, ##L_1##, on top of, ##\epsilon_2## of thickness, ##L_2##. This first stack covers half the plate area. The other stack is ##\epsilon_3## of thickness, ##L=L_1+L_2## covering the other half of the plate area. There are two D's or, what's the same thing, two charge densities.
 
  • #17
Delta2
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Let me see if I get this right: We can use gauss law to prove that the D inside dielectric 1 is equal to D inside dielectric 2 (as long as there are no charges in the interface between these two dielectrics). But there is no way to prove that D inside dielectric 3 is equal to the D of dielectric 1 (or 2). The symmetry present is along the vertical axis and not along the horizontal axis. Did I get this right?
 
  • #18
Paul Colby
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But there is no way to prove that D inside dielectric 3 is equal to the D of dielectric 1 (or 2).
You can't because the physics (and math) says they aren't.

It might help visualization to take a knife split the problem vertically down the middle along the interface between ##\epsilon_1## ##\epsilon_2## and ##\epsilon_3##. Separate the two halves. Connect the two top plates (formerly 1 plate) with a wire. Likewise with the bottom plates. You now have two capacitors with unequal capacitance connected in parallel which electrically identical to the original problem. Each, like before, has a different D.
 
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  • #19
Merlin3189
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I've just been following this thread and am puzzled about a few points:
1- What does ##C_{vac} ## mean? It doesn't seem to be labelled on the diagram. Is it the value that a capacitor with the same A and d, but no dielectric would have?
2- For all the questions, I don't see how you can calculate the answers without the value of A and d.
3-For a) and c) I can't see how you can calculate this without knowing the voltage on the plates. I see some posts refer to V, but I can't see anywhere this is defined.

Does anyone have the answers so that I can check, if I complete this?
 
  • #20
Paul Colby
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I've just been following this thread and am puzzled about a few points:
1) I would assume ##C_{vac}## is the case for ##\epsilon_i= \epsilon_o##.
it better turn out to be
##C_{vac} = \epsilon_o\frac{A}{d} ##
In my posts ##d = L##
3) Assume a voltage V. It cancels from the expression for ##C## anyway.
 
  • #21
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Once you find the electric field in each dielectric, you can use V=-∫Edl to find the potential for each dielectric. Then using C=Q/V, you can find capacitance. The E field through each plate is just 1/(dielectric constant)*(Efield without dielectric). Each capacitance for me came out to a number times (EoA/d) which is the capacitance of a vacuum.
 
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