Old 11-11-2013, 08:19 PM   #1
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Default Dumb electromagnetics question (dielectrics)

Hey y'all. I'm trying to help a student who's stumped on a question... which I can't even answer myself. Am I missing something?

Basically you have a dielectric with varying dielectric constants inside it.

b=1cm
a=5cm (the armatures have a square shape, meaning A=a^2)
1/5 of the dielectric has k1=4
4/5 of the dielectric has k2=6.5

We're looking for the analytic and numerical dielectric constant of this dielectric as a whole.



If you could help, I'd really appreciate it. Is it as simple as taking the average of the k's? We know that C=(k*epsilon)*A/d, where epsilon is the vacuum permittivity. How must we use this? Do we need integrals? Etc.
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Old 11-11-2013, 08:36 PM   #2
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Default Re: Dumb electromagnetics question (dielectrics)

Looking at this...I would assume this is a weighted average as a function of length of the material.

The actual lengths don't matter...just the fraction of total length of the capacitor is necessary.

k(av) = (k1*1/5 + k2*4/5) / 2
k(av) = (4*(1/5) + 6.5*(4/5)) / 2
k(av) = 6

I don't think the actual capacitance is needed here, unless you'll use this dielectric for the total capacitance then sure. If the question is worth more than 4 or 5 marks then it might be more involved, idk
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Old 11-11-2013, 08:39 PM   #3
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Default Re: Dumb electromagnetics question (dielectrics)

I don't know anything about electromagnetics, but if it really is as simple as averaging the k's, I would think you would need to take the distribution of each into account, i.e. a weighted average, since it is not half and half but rather 1/5 and 4/5.

Edit: dang it, ninjad
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Old 11-11-2013, 08:42 PM   #4
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Default Re: Dumb electromagnetics question (dielectrics)

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Originally Posted by Dynam0 View Post
Looking at this...I would assume this is a weighted average as a function of length of the material.

The actual lengths don't matter...just the fraction of total length of the capacitor is necessary.

k(av) = (k1*1/5 + k2*4/5) / 2
k(av) = (4*(1/5) + 6.5*(4/5)) / 2
k(av) = 6

I don't think the actual capacitance is needed here, unless you'll use this dielectric for the total capacitance then sure. If the question is worth more than 4 or 5 marks then it might be more involved, idk
I thought it might have been the case, but there doesn't seem to be any info about this online whatsoever... how strange.

I believe the initial capacitance had to be
C=(epsilon)*A^2/b
=(epislon)*(0.05^2)/0.01
=2.21*10^-12F,

therefore the final capacitance would be 6 times that quantity......?
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Old 11-11-2013, 08:51 PM   #5
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Default Re: Dumb electromagnetics question (dielectrics)

This actually makes more sense than I initially thought it did. Thanks guys, so far.
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Old 11-11-2013, 08:55 PM   #6
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Default Re: Dumb electromagnetics question (dielectrics)

Initial capacitance is before introducing the dielectric, yes? Makes sense since the dielectric of a vacuum is 1.

Yes you should multiply that capacitance by 6 then, assuming the calculation of the dielectric is right
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Old 11-11-2013, 09:08 PM   #7
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Default Re: Dumb electromagnetics question (dielectrics)

I have never studied this shit before *at all* (electrical engineering is not my thing) so here's my completely ignorant attempt (disclaimer here: idk how capacitor shit works or if I am even understanding the model here)

C = k * epsilon_0 * A / d

where
C = capacitance (farad)
A = area (m^2)
d = distance (m)
k = dielectic constant
epsilon_0 = 8.854 * 10^(-12) F/m

For the k1 chunk
C = 4 * 8.854 * 10^(-12) F/m * (.05*.01) m^2 / (.01 m) = 1.7708 * 10^(-12) F

For the k2 chunk
C = 6.5 * 8.854 * 10^(-12) F/m * (.05*.04) m^2 / (.01 m) = 1.15102 * 10^(-11) F


So I'm guessing you just take a weighted average for the whole plate?

C = (1/5)*(1.7708 * 10^(-12) F) + (4/5)*(1.15102 * 10^(-11) F) = 9.56232 * 10^(-12) F

So solve for

k= d*C / (A*epsilon_0) = .01m * 9.56232 * 10^(-12) F / (.05*.05 m^2 * 8.854 * 10^(-12) F/m) = 4.32

(this is probably wrong, I don't know if what I did in this solution was valid)

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Old 11-11-2013, 09:30 PM   #8
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Default Re: Dumb electromagnetics question (dielectrics)

Thanks for your input, rubix... this is raising the question; should we do the weighted average of the capacitances, or of the dielectric constants?
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Old 11-11-2013, 09:34 PM   #9
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Default Re: Dumb electromagnetics question (dielectrics)

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Originally Posted by MarioNintendo
We're looking for the analytic and numerical dielectric constant of this dielectric as a whole.
Uhh... :/
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Old 11-11-2013, 09:38 PM   #10
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Default Re: Dumb electromagnetics question (dielectrics)

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Originally Posted by Reincarnate View Post
I have never studied this shit before *at all* (electrical engineering is not my thing) so here's my completely ignorant attempt (disclaimer here: idk how capacitor shit works or if I am even understanding the model here)

C = k * epsilon_0 * A / d

where
C = capacitance (farad)
A = area (m^2)
d = distance (m)
k = dielectic constant
epsilon_0 = 8.854 * 10^(-12) F/m

For the k1 chunk
C = 4 * 8.854 * 10^(-12) F/m * (.05*.01) m^2 / (.01 m) = 1.7708 * 10^(-12) F

For the k2 chunk
C = 6.5 * 8.854 * 10^(-12) F/m * (.05*.04) m^2 / (.01 m) = 1.15102 * 10^(-11) F

Should be A1 = (.01*.01) and A2 = (.01*.04). Capacitance is directly proportional to both area and the dielectric of the material so the k2 chunk would have a higher capacitance.


So I'm guessing you just take a weighted average for the whole plate?

C = (1/5)*(1.7708 * 10^(-12) F) + (4/5)*(1.15102 * 10^(-11) F) = 9.56232 * 10^(-12) F

So solve for

k= d*C / (A*epsilon_0) = .01m * 9.56232 * 10^(-12) F / (.05*.05 m^2 * 8.854 * 10^(-12) F/m) = 4.32

(this is probably wrong, I don't know if what I did in this solution was valid)


Whether you do it this way or average the dielectric first shouldn't change the final capacitance you compute.
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Old 11-11-2013, 09:40 PM   #11
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Default Re: Dumb electromagnetics question (dielectrics)

I was assuming that A (area) referred to the area of the plate that "corresponded" to whatever's in between with the particular dielectric constant, so like even though that pic is a side view I was assuming something like this:



hence the .05*.01 and .05*.04 in the two equations. Again idk if that's valid.

(Assuming square plates here, too)

Also 1.7708 * 10^(-12) F < 1.15102 * 10^(-11) F implying the second one has higher capacitance yea?

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Old 11-11-2013, 09:44 PM   #12
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Default Re: Dumb electromagnetics question (dielectrics)

True, the ratio of area is still going to be 4:1 for k2:k1. Either way, the k2 chunk shows a lower capacitance in your answer...maybe double check the computations
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Old 11-11-2013, 09:48 PM   #13
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Default Re: Dumb electromagnetics question (dielectrics)

Maybe I am missing the obvious, but doesn't it show a higher (not a lower) capacitance?
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Old 11-11-2013, 09:49 PM   #14
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Default Re: Dumb electromagnetics question (dielectrics)

Missed the exponents being different lmao...sorry for dumb

Logically though, it strikes me as odd that the overall dielectric is closer to 4 than 6.5 in that answer.
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Old 11-11-2013, 09:51 PM   #15
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Default Re: Dumb electromagnetics question (dielectrics)

I agree, Dynam0, that's been bugging me too.

BUT:

Instead of taking a weighted average of the capacitances (since the weighted averaging was, I guess, taken into account via the area partitioning of the initial two equations), if you just add them together lol:

k = .01 * (1.7708 * 10^(-12)+1.15102 * 10^(-11)) / (.05*.05 * 8.854 * 10^(-12) ) = 6

So maybe it works that way instead and it really is that simple, haha

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Old 11-11-2013, 09:54 PM   #16
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Default Re: Dumb electromagnetics question (dielectrics)

Yes! I think that makes sense. Capacitance is the ability to store charge so the value of capacitance in each chunk is additive when representing the whole system.
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Old 11-11-2013, 10:12 PM   #17
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Default Re: Dumb electromagnetics question (dielectrics)

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Originally Posted by Reincarnate View Post
I agree, Dynam0, that's been bugging me too.

BUT:

Instead of taking a weighted average of the capacitances (since the weighted averaging was, I guess, taken into account via the area partitioning of the initial two equations), if you just add them together lol:

k = .01 * (1.7708 * 10^(-12)+1.15102 * 10^(-11)) / (.05*.05 * 8.854 * 10^(-12) ) = 6

So maybe it works that way instead and it really is that simple, haha
Tried to follow, but I get stumped here... What are these numbers?

also, damn that picture was unexpected, lol! You guys work in a professional manner, to say the least.

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Uhh... :/
In other words, we want the algebraic answer and the numerical answer for the capacitance of this dielectric capacitor.
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Old 11-11-2013, 10:18 PM   #18
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Default Re: Dumb electromagnetics question (dielectrics)

Corrected version of my earlier post:

C = k * epsilon_0 * A / d

where
C = capacitance (farad)
A = area (m^2)
d = distance (m)
k = dielectic constant
epsilon_0 = 8.854 * 10^(-12) F/m

For the k1 chunk
C_1 = 4 * 8.854 * 10^(-12) F/m * (.05*.01) m^2 / (.01 m) = 1.7708 * 10^(-12) F

For the k2 chunk
C_2 = 6.5 * 8.854 * 10^(-12) F/m * (.05*.04) m^2 / (.01 m) = 1.15102 * 10^(-11) F

Add the capacitances together to get the value for the whole system:

C_system = C_1 + C_2 = 1.7708 * 10^(-12) F + 1.15102 * 10^(-11) F = 1.3281 * 10^(-11) F

Now we solve for our new k from the original equation (rearranged): k = d*C / (A * epsilon_0)

k_system = .01 m * (1.3281 * 10^(-11) F) / ((.05*.05) m^2 * 8.854 * 10^(-12) F/m) = 6

So in the end, it comes down to the following:

k_system = (1/5)*(4) + (4/5)*(6.5) = 6



Algebraic answer:

k_system = (A_1 / (A_1 + A_2)) * k_1 + (A_2 / (A_1 + A_2)) * k_2

Where A_1 is the area of the plate corresponding to the k_1 stuff and A_2 is the area of the plate corresponding to the k_2 stuff

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Old 11-11-2013, 10:22 PM   #19
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Default Re: Dumb electromagnetics question (dielectrics)

Thanks a bunch you two. I mean it!
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