# Dielectric Influence on Series and Parallel Capacitors

1. Mar 6, 2012

### cellphone

1. The problem statement, all variables and given/known data

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

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').

In Fig. 2, the potential difference between the plates of capacitor 2 _______ when the dielectric is inserted.
In Fig. 2, the charge on capacitor 1 _______ when the dielectric is inserted.
In Fig. 1, the capacitance of capacitor 2 _______ when the dielectric is inserted.
In Fig. 1, the potential energy stored in capacitor 1 _______ when the dielectric is inserted.

2. Relevant equations
ΔV = Q1/C1 = Q2/C2 ( parallel )
Ceq = C1 + C2 ( parallel )
C = Cok

Q is distributed evenly over two capacitors in series.

3. The attempt at a solution

A. Stays the same. ΔV over a capacitance in parallel always remains the same.

B. ΔV = Q1/C1 = Q2/C2, when C2 increases (C = Cok), Q2 will increase as well to compensate. Qnet = Q1 + Q2, this means Q1 must decrease.

C. C = Cok, therefore C increases.

D. 1/Ceq = 1/C1 + 1/C2. when C2 grows larger, 1/C2 grows smaller resulting in a smaller C1. V1 = Q2/(2*C).

I've revised this many times and I still can't see what I'm doing wrong. Any help muchly appreciated, thanks!

Last edited by a moderator: May 5, 2017
2. Mar 6, 2012

### Staff: Mentor

For B, what's the action of the battery with regards to the potential across C1? Can it change?

3. Mar 6, 2012

### cellphone

Well, the battery emits a voltage that is said to be the same across two capacitors in parallel (or so I was taught and have read), so no the voltage across the two capacitors should not change when a dielectric species is introduced. That being said, doesn't that coincide with what I said above? Thanks for the reply, by the way!

4. Mar 6, 2012

### Staff: Mentor

You argued that changing the capacitance of C2 would have an influence on the charge on C1. What determines the charge on C1?

5. Mar 6, 2012

### cellphone

The charge emitted from the battery. When I attempted another answer, I got it right by saying that the charge stays the same. This still confuses me because the voltage shouldn't change, and that being said the charge must change to compensate for the change in capacitance.

6. Mar 6, 2012

### cellphone

Also, if I have the equation ΔV = Q1/C1 = Q2/C2 and CNET = C1 + C2, wouldn't this mean that if even one variable changed (as ΔV is constant), the rest would change?

7. Mar 6, 2012

### Staff: Mentor

Capacitor C1 is not changing value. The voltage supply is not changing value. The voltage supply sets the charge on C1.

8. Mar 6, 2012

### cellphone

So despite the fact that C2 is changing due to a dielectric, C1 wouldn't have to change to compensate?

9. Mar 6, 2012

### Staff: Mentor

That's right. Only C2 has to make changes to alter its charge to match the voltage supply.

10. Mar 6, 2012

### cellphone

Awesome, thank you so much! The way I was looking at it was the same as having an equivalent capacitance. The same way you'd find one to "shrink" a circuit to find the resultant potential differences and charges across capacitors. So in summary, a dielectric only affects the capacitance of it's capacitor and not those around them?

11. Mar 6, 2012

### Staff: Mentor

Correct.

12. Mar 7, 2012

### tiny-tim

welcome to pf!

hi cellphone! welcome to pf!

another way of looking at it (have you done the difference between the D and E fields yet?)

is that the D field, which is measured in units of charge per area (C/m²) is the charge per area (Q/A), and is unaffected by anything else …

so D = Q/A for each capacitor no matter what is between them, and then E = D/ε