# Homework Help: Dielectrics influence on series/parallel

1. Mar 5, 2009

### Dopefish1337

1. The problem statement, all variables and given/known data
Fig. 1 and 2 show a dielectric slab being inserted between the plates of one of two identical capacitors, capacitor 2. Select the correct answer to each of the statements below (enter I for increases', D for decreases', or S for `stays the same').

http://img99.imageshack.us/img99/763/prob45.gif [Broken]

A. In Fig. 1, the potential difference between the plates of capacitor 2 _______ when the dielectric is inserted.

B. In Fig. 1, the charge on capacitor 2 _______ when the dielectric is inserted.

C. In Fig. 2, the capacitance of capacitor 1 _______ when the dielectric is inserted.

D. In Fig. 2, the potential energy stored in capacitor 1 _______ when the dielectric is inserted.

2. Relevant equations

For Capacitors in parallel: Ceq=C1+C2+C3...
In series 1/Ceq=1/C1+1/C2+1/C3...

The introduction of a dielectric increases capcitence to C=kC0

Energy stored on a Capacitor is uc=(1/2)C(deltaVc)2

C=Q/(deltaV)

3. The attempt at a solution

For A, since the capcitors are in series, the charge on each must be constant, as no charge flow could reach the region between the two capacitors. Thus when the dielectric is added to #2, the potential difference across it must go down from deltaV=Q/C. Therefore, D, for decreases.

For B, due to the above reasoning, the charges can't have changed, so the answer would be S, stays the same.

For C, they are in parallel, so the potential difference across each is a constant. When the dielectric is added to #2, it's capacitence would go up, increasing the charge on that one, however, this shouldn't influence the capacitence on #1, so again, S, for stays the same.

For D, the energy is uc=(1/2)C(deltaVc)2, so since the potential difference hasn't changed, and C hasn't changed, this would also be S, stays the same.

I feel reasonably confident that C and D should have the same answer, so I suspect if one of those two is wrong, the other is wrong in the same way. (I.E. if C is really I, then so is D.)

I know theres at least one mistake in my reasoning somewhere, I'm just not sure where.

Last edited by a moderator: May 4, 2017
2. Mar 6, 2009

### alphysicist

Hi Dopefish1337,

Just looking at Figure 1, I believe you are incorrect in saying that the charges on the capacitors cannot change. These capacitors in series are charged because charges from the "top" plate of capacitor #2 move to the "bottom" plate of capacitor #1, and so the capacitor charge can change without crossing through the capacitor gap itself. (Also, charges move from the "top" of capacitor 1 to the "bottom" of #2.)

I would suggest determining what happens to the total capacitance (of both capacitors) when the dielectric is inserted, and what that means for the charge on each capacitor since they are in series.

Last edited by a moderator: May 4, 2017
3. Mar 6, 2009

### Dopefish1337

Does my reasoning for the questions pertaining to figure two seem correct?

4. Mar 6, 2009

### Dopefish1337

Since in the first figure they're attached to a battery, does that imply they all have the same potential difference, making my answer to a A. S, does not change?

And for B... Sticking in some test values for the capacitences, voltage, and kappa respectively, it seems to show that the total charge on the capcitors decreases when the dielectric is added. So, would that mean the answer to B is D, it'd decrease?

Where does that charge go? Does it get absorbed into the battery?

5. Mar 6, 2009

### alphysicist

No, the total potential difference across both capacitors together does not change, but how much is across each one will change.

So before you had the dielectric inserted, each capacitor had a potential difference of V/2 (since they are identical). However, when the dielectric is inserted this is no longer true. The sum of the voltages will still equal the battery voltage V, but they will no longer be identical.

Can you show how you got that the total charge decreases?

6. Mar 6, 2009

### Dopefish1337

To determine the result of Q decreasing, I let C1=C2=1 F, V=1 V, and K=2, since the result of this should result for any given values of the above (outside perhaps a K<1, but those don't exist).

So, without the dielectric, the Ceq would be (1/1+1/1)-1, or 0.5.

Since Ceq=Q/V, with V=1, that leaves as hypothetical charge of 0.5.

With the dielectric, the new Ceq would become (1/1+1/2)-1, or 2/3. Again, with a assumed V of 1, that leaves the resultant Q to increase from 0.5 to 0.67

I believe the first time I must've forgoten to take the reciprocal, leaving my initial Ceq as 2, rather than 1/2.

This 'feels' more right to me, because dielectrics increase capacitence, which has been described as the ability to hold charge to me, so this result seems much better.

I'm a little shakey as to how to approach A though...I feel as theres something very obvious right in front of me, but I can't seem to put my finger on it.

edit: Actually, could I use a similar process and the knowledge that V1+V2=V, to show the potential difference on V2 must decrease? (without wishing to type out the math, getting 1/2 normally, and 1/3 with the dielectric, based on my assumed numbers.)

Last edited: Mar 6, 2009
7. Mar 6, 2009

### alphysicist

That's right; so the total charge will increase.

Continue on with your numbers. With those numbers, you found the total charge on the equivalent capacitance (with the dielectric in place) to be Q=2/3. Since the two capacitors are in series, what is the charge on each one of them? Then, once you know the charge on a capacitor and the capacitance, what is the voltage across C2? Compare that to your original voltage across C2; what do you get?

8. Mar 6, 2009

### Dopefish1337

See my edit for what I got - The voltage must decrease.

Thanks a lot by the way.

Before I go off using up my last few tries, do my initial conclusions about the stuff in figure two seem correct, or did I muddle things up there too?

9. Mar 6, 2009

### alphysicist

That sounds right to me.

You're welcome!