Parallel-plate capacitor: Two dielectric materials

[phyzmatix]

[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

$$C = \frac{\kappa \epsilon_{0} A}{d}$$

$$\frac {1}{C_{tot}} = \frac {1}{C_{left}} + \frac {1}{C_{right}}$$

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:

$$C_{left} = 3.10 pF$$
$$C_{right} = 5.31 pF$$

$$C_{tot} = 1.96 pF$$

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

Cheers

phyz

 

[Phizyk]

This is a serial connection of capacitors so you must add right and left halves capacity:
$$C=C_{right}+C_{left}$$

 

[phyzmatix]

Thanks for your reply...however, could you please explain why this setup acts as capacitors in series?

 

[Phizyk]

I apologize to bad equation $$C=C_{1}+C_{2}$$
It's wrong... You have a good solution, because
$$V=\frac{q}{C}$$
and $$V_{1}=\frac{q}{C_{1}}$$ $$V_{2}=\frac{q}{C_{2}}$$
and $$V=V_{1}+V_{2}$$
so $$\frac{1}{C}=\frac{1}{C_{1}}+\frac{1}{C_{2}}$$

 

[phyzmatix]

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...

 

[tiny-tim]

$$C_{left} = 3.10 pF$$
$$C_{right} = 5.31 pF$$

Hi phyzmatix!

Thanks for the PM …

Did you divide by 2 instead of multiply?

 

[phyzmatix]

Hi phyzmatix!

Thanks for the PM …

Did you divide by 2 instead of multiply?

Thanks for joining me here tiny-tim

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 $$C_{left}$$ (calculated with K1 and A/2) and the right half has capacitance $$C_{right}$$ (calculated with K2 and A/2) but how to combine the two values into $$C_{tot}$$?

(I'm not sure my assumption that the two can be treated as separate parallel capacitors is correct)

 

[tiny-tim]

Hi phyzmatix!

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? )

 

[phyzmatix]

Hi phyzmatix!

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? )

*DOH!*

I feel like such a tw@t

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!

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