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Parallel-plate capacitor: Two dielectric materials

  • Thread starter phyzmatix
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  • #1
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[SOLVED] Parallel-plate capacitor: Two dielectric materials

Homework Statement



A parallel-plate capacitor with area A = 5.56 cm^2 and separation d = 5.56 mm has the left half of the gap filled with material of dielectric constant K1 = 7.00 and the right half filled with material of dielectric constant K2 = 12.0. What is the capacitance?


Homework Equations



[tex]C = \frac{\kappa \epsilon_{0} A}{d} [/tex]

[tex] \frac {1}{C_{tot}} = \frac {1}{C_{left}} + \frac {1}{C_{right}}[/tex]

The Attempt at a Solution



Since it's one capacitor (two plates) and since the dielectric materials are next to each other (as opposed to stacked) I approached the problem as if the two halves of the capacitor were separate parallel capacitors and calculated:

[tex]C_{left} = 3.10 pF [/tex]
[tex]C_{right} = 5.31 pF [/tex]

[tex]C_{tot} = 1.96 pF [/tex]

But this is not amongst the given possible answers. Where did I go wrong?

Cheers

phyz
 

Answers and Replies

  • #2
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This is a serial connection of capacitors so you must add right and left halves capacity:
[tex]C=C_{right}+C_{left}[/tex]
 
Last edited:
  • #3
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Thanks for your reply...however, could you please explain why this setup acts as capacitors in series?
 
  • #4
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I apologize to bad equation [tex]C=C_{1}+C_{2}[/tex]
It's wrong... You have a good solution, because
[tex]V=\frac{q}{C}[/tex]
and [tex]V_{1}=\frac{q}{C_{1}}[/tex] [tex]V_{2}=\frac{q}{C_{2}}[/tex]
and [tex]V=V_{1}+V_{2}[/tex]
so [tex]\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex]
 
  • #5
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But this is not amongst the given possible answers.
I appreciate your attempt at helping me Phizyk, but unfortunately we're no closer to understanding this problem than before...
 
  • #6
tiny-tim
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[tex]C_{left} = 3.10 pF [/tex]
[tex]C_{right} = 5.31 pF [/tex]
Hi phyzmatix! :smile:

Thanks for the PM …

Did you divide by 2 instead of multiply?
 
  • #7
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Hi phyzmatix! :smile:

Thanks for the PM …

Did you divide by 2 instead of multiply?
Thanks for joining me here tiny-tim :smile:

Yes, I divided the total area given by 2 (I wish I had the means to show you the figure given with the question, on the figure they show that the left side is A/2 and the right side is also A/2)

So my understanding is that the left half of the parallel-plate capacitor has capacitance [tex]C_{left}[/tex] (calculated with K1 and A/2) and the right half has capacitance [tex]C_{right}[/tex] (calculated with K2 and A/2) but how to combine the two values into [tex]C_{tot}[/tex]?

(I'm not sure my assumption that the two can be treated as separate parallel capacitors is correct)
 
  • #8
tiny-tim
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Hi phyzmatix! :smile:

Well, they're side-by-side, not one-after-the-other, so they're "in parallel" rather than "in series" (plus, the question calls it a parallel-plate capacitor), so I think you just add them.

Isn't 8.41 one of the given answers?

(if not, what are the given answers? :wink:)
 
  • #9
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Hi phyzmatix! :smile:

Well, they're side-by-side, not one-after-the-other, so they're "in parallel" rather than "in series" (plus, the question calls it a parallel-plate capacitor), so I think you just add them.

Isn't 8.41 one of the given answers?

(if not, what are the given answers? :wink:)
*DOH!!!*

I feel like such a tw@t :redface:

No wonder I didn't get the right answer, I was using the equation for capacitors in series...

*banging head against wall*

Cheers Tim! You legend! :biggrin:

(oh, by the by, 8.41 is indeed amongst the possibilities :wink:)
 

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