# Dielectric Influence on Series and Parallel Capacitors

• cellphone
In summary: Correct.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 :wink:
cellphone

## Homework Statement

http://img99.imageshack.us/img99/763/prob45.gif

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.

## Homework Equations

ΔV = Q1/C1 = Q2/C2 ( parallel )
Ceq = C1 + C2 ( parallel )
C = Cok

Q is distributed evenly over two capacitors in series.

## 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!

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

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!

cellphone said:
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!

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

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.

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?

cellphone said:
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.

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

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

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

cellphone said:
So despite the fact that C2 is changing due to a dielectric, C1 wouldn't have to change to compensate?
That's right. Only C2 has to make changes to alter its charge to match the voltage supply.

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?

cellphone said:
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?

Correct.

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

(btw, are you ok now on your other thread?)

## 1. How does the presence of a dielectric material affect the capacitance of a series capacitor arrangement?

When a dielectric material is inserted between the plates of a series capacitor, the overall capacitance increases. This is because the dielectric material has a higher permittivity than air, which means it can store more electric charge. The electric field between the plates is also reduced, resulting in an increase in capacitance.

## 2. Is the effect of a dielectric material the same for a parallel capacitor configuration?

No, the effect of a dielectric material on a parallel capacitor configuration is different. In this case, the overall capacitance decreases when a dielectric material is inserted between the plates. This is because the dielectric material reduces the effective area between the plates, which decreases the capacitance. However, the electric field between the plates is also reduced, resulting in a decrease in capacitance.

## 3. How does the type of dielectric material used affect the overall capacitance?

The type of dielectric material used can greatly affect the overall capacitance. Different materials have different permittivity values, which determine the extent of their influence on the capacitance. Materials with higher permittivity, such as ceramic or paper, will have a greater impact on the overall capacitance compared to materials with lower permittivity, such as air or vacuum.

## 4. Can the presence of a dielectric material affect the voltage distribution in a capacitor arrangement?

Yes, the presence of a dielectric material can affect the voltage distribution in a capacitor arrangement. When a dielectric material is inserted between the plates, the electric field is reduced, and the voltage is no longer distributed evenly across the plates. The voltage will be higher on the plates closer to the source and lower on the plates closer to the dielectric material.

## 5. How does the distance between the plates of a capacitor affect the influence of a dielectric material?

The distance between the plates of a capacitor plays a significant role in the influence of a dielectric material. As the distance between the plates decreases, the effect of the dielectric material on the capacitance increases. This is because the electric field between the plates is stronger, meaning the dielectric material has a greater impact on reducing the electric field and increasing the capacitance. Conversely, as the distance between the plates increases, the effect of the dielectric material decreases.

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