Calculating charge outside capacitor plates

[Shaybay92]

Homework Statement

A capacitor made of two parallel uniformly charged circular metal disks carries a charge +Q and -Q on the inner surfaces of the the plates and very small amounts of charge +q and -q on the outer surfaces of the plates. Each plate has a radius R and thickness t, and the gap distance between the plates is s. How much charge q is on the outside surface of the positive disk, in terms of Q?

The Attempt at a Solution

I was going to try and find the electric field at the fringe of the plate by summing up all the contributions of the charges, by treating the faces of the plates as different charged plates.

Fringe field = E+Q + E-Q + E(-fringe) + E(+fringe)
Fringe field - E+Q - E-Q = E(-fringe) + E(+fringe)

I know that magnitude of +q and -q are the same so I would rearrange to find q. Have I made correct assumptions?

 

[mark726]

Hello there,

Your approach seems reasonable, but there are a few things to consider. First, when summing up the contributions of the charges, you should take into account the direction of the electric field, as it will affect the net field at the fringe of the plate.

Secondly, you should also consider the distance of each charge from the fringe of the plate, as this will also affect the strength of the electric field. So your equation for the fringe field should include the distance from the charges to the fringe.

Finally, when rearranging to find q, you should also consider the sign of the charges. Since the charge on the outside surface of the positive disk is +q, the electric field at that point should be pointing away from the disk. This means that the electric field from the positive charges (+Q and +q) should be subtracted from the electric field from the negative charges (-Q and -q).

Overall, your approach is on the right track, but make sure to consider the direction and distance of the electric field, as well as the signs of the charges, in your calculations. Let me know if you have any further questions. Good luck!