Question on capacitance and electric fields

[daviddeakin]

Consider a thin parallel plate capacitor of an area of of 2.0x10^-4 metres squared, with
a thickness of 1.2x10^-4 metres. The top and bottom thirds of the capacitor are filled with a dielectric material with a relative dielectric permittivity εr1=4, and the central third, with another material with εr2=8 (so it's a triple layer sandwich) .The voltage on the capacitor is 3V.

Calculate the electric flux densities and the electric field magnitudes in all three parts of the device. State any assumptions used.

Answer:
We can treat this as three capacitors in series, each with a plate separation of 4x10^-5 m. The upper and lower capacitors are:
C=εA/d = 4 x 8.85x10^-12 / 4x10^-5 = 117pF
The middle capacitor is:
C=εA/d = 8 x 8.85x10^-12 / 4x10^-5 = 354pF

But I don't know how to complete the question. For a single capacitor I would find the charge: Q=CV, then find the electric field from: E = D/ε = Q/Aε

But in this case I don't know what the voltage on each capacitor is. Is it simply one third of the voltage across the whole device? Also, I must be going about this in the wrong way, because the NEXT part of the question is:

Calculate the charge on the plates and the capacitance of the device.

The total capacitance is clearly:
C = [1/C1 + 1/C2 + 1/C3]^-1 = 50.2pF

But the way I'm attempting it already gives the charge before the second part of the question!

(Also, is there a way to do superscripts on this forum?)

 

[gneill]

Hi daviddeakin, Welcome to Physics Forums.

Consider a thin parallel plate capacitor of an area of of 2.0x10^-4 metres squared, with
a thickness of 1.2x10^-4 metres. The top and bottom thirds of the capacitor are filled with a dielectric material with a relative dielectric permittivity εr1=4, and the central third, with another material with εr2=8 (so it's a triple layer sandwich) .The voltage on the capacitor is 3V.

Calculate the electric flux densities and the electric field magnitudes in all three parts of the device. State any assumptions used.

Answer:
We can treat this as three capacitors in series, each with a plate separation of 4x10^-5 m. The upper and lower capacitors are:
C=εA/d = 4 x 8.85x10^-12 / 4x10^-5 = 117pF
The middle capacitor is:
C=εA/d = 8 x 8.85x10^-12 / 4x10^-5 = 354pF

I think you forgot to include the plate area in your calculations above. I see the numbers for permittivity and distance, but not the area.

But I don't know how to complete the question. For a single capacitor I would find the charge: Q=CV, then find the electric field from: E = D/ε = Q/Aε

But in this case I don't know what the voltage on each capacitor is. Is it simply one third of the voltage across the whole device?

Since the capacitors are in series, what can you say about the charge on each?

Also, I must be going about this in the wrong way, because the NEXT part of the question is:

Calculate the charge on the plates and the capacitance of the device.

The total capacitance is clearly:
C = [1/C1 + 1/C2 + 1/C3]^-1 = 50.2pF

But the way I'm attempting it already gives the charge before the second part of the question!

(Also, is there a way to do superscripts on this forum?)

Try the X² icon in the the edit window header, or use the LaTex syntax (you'll need to learn about LaTex embedding tags).

The person who created the problem may have had a particular strategy in mind, but so long as you get to the correct solution by valid steps don't worry about the order that the parts get answered.

 

[daviddeakin]

Since the capacitors are in series, what can you say about the charge on each?

Hi Gneill,

I have difficulty getting to grips with charge! If the top plate has some positive charge, then the bottom plate must have some equal negative charge. But the ones in between I'm not sure about! Do they have similar charge, alternating in sign, or simply zero charge?

If it is zero, then the electric field in the central capacitor would be zero, which sounds wrong somehow...

The plate area was 2.0x10^-4 metres squared, BTW.

 

[gneill]

Current is always the same for all components connected in series. So if charge Q is moved onto the top plate of the stack, charge Q must move off of the next plate in line and onto the plate following that, and so on down the line. The net result is that all capacitors in a series connection end up with the same charge (+Q on one plate, -Q on the other).

 

[daviddeakin]

The net result is that all capacitors in a series connection end up with the same charge (+Q on one plate, -Q on the other).

Intriguing! So the top and bottom capacitors are:
C=εA/d = 4 x 8.85x10^-12 x 2.0x10^-4 / 4x10^-5 = 177pF
The middle capacitor is:
C=εA/d = 8 x 8.85x10^-12 x 2.0x10^-4 / 4x10^-5 = 354pF

The total capacitance is: 70.8pF

Charge on the top and bottom plates is:
Q = CV = 70.8pF x 3V = 212.4pC
which must now be the same for all of the capacitors.

The electric fields in the top and bottom capacitors are:
E = Q/εA = 212.4pC / (4 x 8.85x10^^-12 x 2.0x10^-4 ) = 30kV/m
And for the middle capacitor:
E = Q/εA = 212.4pC / (8 x 8.85x10^^-12 x 2.0x10^-4 ) = 15kV/m

The flux density is assumed uniform and will be the same for all the capacitors:
D = Q/A = 212.4pC / 2.0x10^-4 = 1.1 uC/m²

Thanks!

 

[gneill]

Intriguing! So the top and bottom capacitors are:
C=εA/d = 4 x 8.85x10^-12 x 2.0x10^-4 / 4x10^-5 = 177pF
The middle capacitor is:
C=εA/d = 8 x 8.85x10^-12 x 2.0x10^-4 / 4x10^-5 = 354pF

The total capacitance is: 70.8pF

Charge on the top and bottom plates is:
Q = CV = 70.8pF x 3V = 212.4pC
which must now be the same for all of the capacitors.

The electric fields in the top and bottom capacitors are:
E = Q/εA = 212.4pC / (4 x 8.85x10^^-12 x 2.0x10^-4 ) = 30kV/m
And for the middle capacitor:
E = Q/εA = 212.4pC / (8 x 8.85x10^^-12 x 2.0x10^-4 ) = 15kV/m

The flux density is assumed uniform and will be the same for all the capacitors:
D = Q/A = 212.4pC / 2.0x10^-4 = 1.1 uC/m²

Looks good, except you might want to check whether they want charge density (Q/A) or electric flux density ($\Phi_E$) for that last bit. They are different beasts.

 

[daviddeakin]

you might want to check whether they want charge density (Q/A) or electric flux density ($\Phi_E$) for that last bit. They are different beasts.

Is ($\Phi$) the flux in coulombs? If so then ($\Phi_E$) would be the force between the plates wouldn't it?

 

[gneill]

Is ($\Phi$) the flux in coulombs? If so then ($\Phi_E$) would be the force between the plates wouldn't it?

Charge is given in coulombs. Flux is in volt-meters (Vm).

Remember Gauss' Law where the electric flux crossing the boundary of an enclosed space ("Gaussian Surface") is summed up in order to determine the enclosed charge?

$$\Phi_E = \oint_s \! E \cdot dA = \frac{Q_s}{\epsilon_r\epsilon_o}$$

 

[daviddeakin]

Charge is given in coulombs. Flux is in volt-meters (Vm).

Remember Gauss' Law where the electric flux crossing the boundary of an enclosed space ("Gaussian Surface") is summed up in order to determine the enclosed charge?

$$\Phi_E = \oint_s \! E \cdot dA = \frac{Q_s}{\epsilon_r\epsilon_o}$$

Hmm, I've been learning from a book which says:

"Gauss' law states that the flux through any closed surface is equal to the charge enclosed by that surface."

So I have been using flux in coulombs... What am I missing?

 

[gneill]

Hmm, I've been learning from a book which says:

"Gauss' law states that the flux through any closed surface is equal to the charge enclosed by that surface."

So I have been using flux in coulombs... What am I missing?

I think you're missing the bit at the end of that sentence that goes "...divided by the permittivity."

Charge is in Coulombs. Flux is in Volt-meters. The idea is that flux is what charges produce that describes the electric field. It is similar in concept to magnetic flux lines.

You have a couple of choices of method for calculating the flux here. The field is constant over the area of the plates of the capacitors so you can multiply E*A, which is the same as the performing the surface integral I showed above, or you can divide the plate charge by the permittivity for the given capacitor.

 

[daviddeakin]

I think you're missing the bit at the end of that sentence that goes "...divided by the permittivity."

There's definitely nothing else at the end of sentence! The whole book (and indeed my lecture course) follows the same lines- flux is in coulombs. See Wiki, for example:

http://en.wikipedia.org/wiki/Statcoulomb
"An electric flux (specifically, a flux of the electric displacement field D) has units of charge: statC in cgs and coulombs in SI."

Perhaps physicists and electronic-ists have different conventions...

 

[gneill]

There's definitely nothing else at the end of sentence! The whole book (and indeed my lecture course) follows the same lines- flux is in coulombs. See Wiki, for example:

http://en.wikipedia.org/wiki/Statcoulomb
"An electric flux (specifically, a flux of the electric displacement field D) has units of charge: statC in cgs and coulombs in SI."

Perhaps physicists and electronic-ists have different conventions...

Ah. Statcoulombs. That's a unit system that handles things a bit differently. That's not the system that typically used for most courses. I rarely run across it at all.

Note the statement of Gauss' Law here: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gaulaw.html

Also here: http://en.wikipedia.org/wiki/Gauss%27s_law

 

[daviddeakin]

Ah. Statcoulombs. That's a unit system that handles things a bit differently.

I don't think it's statcoulombs- they're for the cgs system. Note that it says the flux is in coulombs for the SI system. I can see this is going to keep me awake for a while.EDIT:
I emailed my lecturer about it, and he said:

"It's the same. Intergral of _D_ is charge and integral of _E_ is charge/epsilon.
Remember that _D_= epsilon_E_ by definition."

I can't decide if that helps...