Capacitor Problem

jena

Hi,

Question:

An uncharged capacitor is connected to a 24.0-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin is then inserted between the plates. what will now be the voltage between the plates?

I know that as long as the capacitor stays connect the battery, the voltage stays constant and equal to the battery, since its disconnected the charge will have nowhere to go. I also know from my reading that the capacitance increases because of the dielectric(paraffin), but the voltage decrease because of it.

After this I'm lost. Do I need the dielectric strength of the paraffin to help in this problem.

Thank You

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teclo

jena said:
Hi,

Question:

An uncharged capacitor is connected to a 24.0-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin is then inserted between the plates. what will now be the voltage between the plates?

I know that as long as the capacitor stays connect the battery, the voltage stays constant and equal to the battery, since its disconnected the charge will have nowhere to go. I also know from my reading that the capacitance increases because of the dielectric(paraffin), but the voltage decrease because of it.

After this I'm lost. Do I need the dielectric strength of the paraffin to help in this problem.

Thank You

it's been awhile since i've done something like this, but it seems like you'll have to know the width of the insulator sheet. what happens to the molecules in the insulator when they're put in between the plates. draw it out and draw the charge configurations. like i said i could be wrong, but i think this problem only depends on the geometry of the sheet.

Jelfish

Think of it this way: There are field lines that run from the positive plate to the negative plate of the capacitor. Because the paraffin is an insulator, the electrons on the atoms will not so easily move off the atoms and create a current. The result is that the atoms of the paraffin start to seperate slightly with the postive nucleous pulling toward the negative plate and the electron cloud pulling toward the positive plate. This seperation of the atoms creates an E-field of their own which will be in the opposite direction of the plates, effectively cancelling out some of it. You need to find the equation that gives you the electric field in the presence of a dielectric. Then there is a specific formula the relates the electric field between the plates with the potential difference.

Hope that helps.

jena

Hi,

Thank you for the help of the previous two responses, but I'm still stuck. I don't have any dimension of the paraffin slab, all that I have is the dielectric strength as well as its constant. For this type of problem could I find the capacitance?? I'm not sure

Thank You

jena

Hi,

I found out how to get the answer

V= V(o)/k

where

V(o)=the orginal voltage

and

k= dielectric constant of the paraffin

Thank You

lightgrav

Homework Helper
You know why the distance isn't needed?
Becuase the original Voltage was E_o*d.
The dielectric reduces E_o -> E_o /k ,
so V_o -> V_o /k since d is the same.

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