Parallel-plate capacitor: Two dielectric materials

In summary, the parallel-plate capacitor 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. The capacitance is 1.96 pF.
  • #1
phyzmatix
313
0
[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
 
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  • #2
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
Thanks for your reply...however, could you please explain why this setup acts as capacitors in series?
 
  • #4
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
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
phyzmatix said:
[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
tiny-tim said:
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
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
tiny-tim said:
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:)
 

1. What is a parallel-plate capacitor?

A parallel-plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a dielectric material. It is used to store electrical energy by creating an electric field between the plates.

2. How does a parallel-plate capacitor work?

When a voltage is applied to the capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which stores energy in the form of electric potential. The dielectric material between the plates helps to increase the capacitance and store more energy.

3. What are dielectric materials and why are they used in a parallel-plate capacitor?

Dielectric materials are insulating materials that do not conduct electricity. They are used in a parallel-plate capacitor to increase the capacitance, which is a measure of the capacitor's ability to store charge. The dielectric material reduces the electric field between the plates, allowing for more charge to be stored.

4. What are some common types of dielectric materials used in a parallel-plate capacitor?

Some common types of dielectric materials used in a parallel-plate capacitor include air, paper, mica, ceramic, and plastic. The choice of dielectric material depends on the desired capacitance, voltage rating, and other factors.

5. How is the capacitance of a parallel-plate capacitor calculated?

The capacitance of a parallel-plate capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the area of the plates, and d is the distance between the plates. This formula shows that the capacitance is directly proportional to the area of the plates and the permittivity of the dielectric material, and inversely proportional to the distance between the plates.

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