11-11-2013, 08:19 PM | #1 |
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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. Last edited by MarioNintendo; 11-11-2013 at 08:48 PM.. |
11-11-2013, 08:36 PM | #2 |
The Dominator
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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 |
11-11-2013, 08:39 PM | #3 |
Kawaii Desu Ne?
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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 |
11-11-2013, 08:42 PM | #4 | |
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Re: Dumb electromagnetics question (dielectrics)
Quote:
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......? Last edited by MarioNintendo; 11-11-2013 at 08:50 PM.. |
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11-11-2013, 08:51 PM | #5 |
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Re: Dumb electromagnetics question (dielectrics)
This actually makes more sense than I initially thought it did. Thanks guys, so far.
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11-11-2013, 08:55 PM | #6 |
The Dominator
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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 |
11-11-2013, 09:08 PM | #7 |
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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) Last edited by Reincarnate; 11-11-2013 at 09:17 PM.. |
11-11-2013, 09:30 PM | #8 |
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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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11-11-2013, 09:34 PM | #9 | |
Fractals!
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Re: Dumb electromagnetics question (dielectrics)
Quote:
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11-11-2013, 09:38 PM | #10 | |
The Dominator
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Re: Dumb electromagnetics question (dielectrics)
Quote:
Whether you do it this way or average the dielectric first shouldn't change the final capacitance you compute. |
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11-11-2013, 09:40 PM | #11 |
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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? Last edited by Reincarnate; 11-11-2013 at 09:43 PM.. |
11-11-2013, 09:44 PM | #12 |
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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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11-11-2013, 09:48 PM | #13 |
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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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11-11-2013, 09:49 PM | #14 |
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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. |
11-11-2013, 09:51 PM | #15 |
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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 Last edited by Reincarnate; 11-11-2013 at 09:54 PM.. |
11-11-2013, 09:54 PM | #16 |
The Dominator
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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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11-11-2013, 10:12 PM | #17 | |
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Re: Dumb electromagnetics question (dielectrics)
Quote:
also, damn that picture was unexpected, lol! You guys work in a professional manner, to say the least. In other words, we want the algebraic answer and the numerical answer for the capacitance of this dielectric capacitor. Last edited by MarioNintendo; 11-11-2013 at 10:14 PM.. |
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11-11-2013, 10:18 PM | #18 |
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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 Last edited by Reincarnate; 11-11-2013 at 10:46 PM.. |
11-11-2013, 10:22 PM | #19 |
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Re: Dumb electromagnetics question (dielectrics)
Thanks a bunch you two. I mean it!
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