Dumb electromagnetics question (dielectrics)

[MarioNintendo]

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.

[Dynam0]

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

[benguino]

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

[MarioNintendo]

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

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......?

[MarioNintendo]

Re: Dumb electromagnetics question (dielectrics)

This actually makes more sense than I initially thought it did. Thanks guys, so far.

[Dynam0]

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

[Reincarnate]

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)

[MarioNintendo]

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?

[igotrhythm]

Re: Dumb electromagnetics question (dielectrics)

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

Uhh... :/

[Dynam0]

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

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.

[Reincarnate]

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?

[Dynam0]

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

[Reincarnate]

Re: Dumb electromagnetics question (dielectrics)

Maybe I am missing the obvious, but doesn't it show a higher (not a lower) capacitance?

[Dynam0]

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.

[Reincarnate]

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