How would this capacitor configuration work?

[azure kitsune]

Assume you have two parallel plates with charges +q and -q.

Now, you fill the space in between with two different dielectrics so that if you were to move straight from one plate to the other, you would be in the same dielectric. (In other words, the two dielectrics do not form "layers.")

My question is: What happens to the charges on the plates?

Do they stay uniformly distributed? This would mean the potential difference across the capacitor is different at different locations. Wouldn't this make the electric field inside the capacitor non-conservative?

Do the charges move reposition themselves so that the potential difference remains constant? If this were true, wouldn't the charges have a tendency to go back to uniform distribution because the closer-packed ones would repel each other more?

If the capacitor is connected to an EMF source with constant voltage, then I believe the charges would reposition themselves to maintain the same potential drop. But what if there is no current?

 

[kcdodd]

The conductors will always be equipotential, due to E = 0 inside the conductor. The dielectrics are polarized in the direction of E, which means there is an effective charge density next to each conductor. This allows the charge inside the conductor to be re-arranged.

 

[azure kitsune]

Thank you for explaining! I think I understand. Is the following statement correct?:

For a charged conducting plate by itself, the charge would be uniformly distributed on the two surfaces because there is no other way for E to be 0 inside. For a capacitor with two dielectrics, the charge is not uniformly distributed, so the electric field inside the plate due to the charges on the surface alone is not zero. However, the net electric field inside the conductor (which includes E from the polarized dielectrics) is zero.

 

[kcdodd]

It's not really safe to assume charge is distributed uniformly, even for isolated conductor. Except perhaps spheres or infinite plates. You usually start with the assumption of a constant voltage on the conductor, and then solve for the charge density.

 

[sardar]

The capacitor configuration described would work by storing electrical energy in the form of electric potential energy between the two charged plates. The presence of the dielectric material between the plates would affect the electric field and therefore the potential difference between the plates.

In this configuration, the charges on the plates will redistribute themselves to maintain a constant potential difference between the plates, as the electric field is affected by the dielectric material. This means that the charges will not remain uniformly distributed, but will instead accumulate more on one side of the plates depending on the dielectric constant of the material.

The electric field inside the capacitor may not be conservative in this case, as the potential difference is different at different locations. However, this does not affect the overall functioning of the capacitor as it is still able to store energy.

In the absence of an external EMF source, the charges on the plates will eventually redistribute themselves to maintain a uniform distribution. This is due to the repulsive forces between like charges. However, if the capacitor is connected to a constant voltage source, the charges will continue to maintain their redistributed positions.

In summary, the capacitor configuration described will work by storing electrical energy between two charged plates, with the presence of dielectric material affecting the potential difference and charge distribution. The charges will redistribute themselves to maintain a constant potential difference, and in the absence of an external EMF source, will eventually return to a uniform distribution.