# Capacitor with partially inserted Dielectric

1. Nov 16, 2004

### joshanders_84

This is one of my homework problems that I am stumped with:

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.

A) A voltmeter reads V_1 when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads V_2. What is the dielectric constant of this material?

I know the answer to this: it is V_1/V_2 (as the definition of K).

B) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

I don't understand how to answer this one. I have tried going into the definition of capacitance with and without a dielectric, and came up with V_1/(k/3), but that is not the right answer. I don't understand how to figure this out...any and all help is much appreciated. Thank you

2. Nov 16, 2004

### dduardo

Staff Emeritus
Use the equation C = Q/V = (epsilon*S)/d

In the case when you only have free space what is V1?

For the case where you have 1/3 dieletric and 2/3 free space you can split the problem into two capacitors in parallel.

3. Nov 16, 2004

### joshanders_84

In that case, shouldn't it be (1/3)V_2 + (2/3)V_1? I have tried that but it is incorrect and I really just don't seem to get it.

4. Nov 16, 2004

### dduardo

Staff Emeritus
Um, no. I believe the answer should only have a V1.

5. Nov 16, 2004

### gnome

Remember that the total charge doesn't change, but the charge density in the region with the dielectric will be different from the region without. The charges must end up distributed so that the voltage is the same everywhere on the plates.

Ed: Solve for Ceq as dduardo suggested. Then find an expression for Q in terms of V1 and use the fact that Ceq = Q/Vf

PS: I wouldn't stake my life on it, but my answer has both V1 and V2 in it.

Last edited: Nov 16, 2004
6. Nov 17, 2004

### joshanders_84

But in this case, Ceq is equal to what? I mean how can I solve for that?

7. Nov 17, 2004

### gnome

In effect you have 2 capacitors in parallel, so use
$$\frac{1}{C_{eq}}=\frac{1}{C_1} + \frac{1}{C_2}$$

You can express C1 as a function of &epsilon;0, d, A, and V1 and C2 as a function of &epsilon;0, d, A, and V2.

Put all of that together, along with the knowledge that Q is a constant that you can express as a function of &epsilon;0, d, A, and V1 or V2, will give you the answer.

8. Nov 17, 2004

### dduardo

Staff Emeritus
gnome, the equation you gave is for capacitors in series. You just add the capacitor values in parallel: C1+C2+...Cn

9. Nov 17, 2004

### gnome

Hmmm...resistance, capacitance... can't keep 'em straight after 1 year :grumpy:

Yes, you're absolutely right. Thanks.

So,
$$C_{eq}={C_1} + C_2$$

gives me
$$V_f = \frac{3V_1V_2}{V_1 + 2V_2}$$

Do you agree?