Unsymmetrically Charged Plate Capacitor, electrostatics

[guggikofod]

I think I have a problem with some fundamentals, please correct me.

Problem:

Given: A vacuum capacitor with electrodes of area A and distance d is initially uncharged. The plates are disconnected from ground, and are not connected to each other with a wire.

Using a rod, Plate 1 is charged to a charge of Q1. Likewise, the other plate is charged to Q2. These charges are unequal, not Q1 = - Q2. Let's set Q1 > Q2.

What is the voltage potential drop between the two plates (in terms of the charges on the plates)?

I have been using Gauss boxes, which results in consistent equations, however, I couldn't solve it. Any suggestions?

Thanks

[bm1957]

Remember that:

C = q/V, where q is Q1-Q2

and note that for an ideal capacitor (which a vacuum capacitor is):

C = (epsilon 0)A/d, where epsilon 0 is the permittivity of free space.

A little algebra should yield V = ... you will get an expression in terms of A, d and q

Post back if you need any more help.

[Enthalpy]

You can decompose mathematically your two charges in a mean value, and a difference to the mean value - this difference is then symmetrical.

As long as capacitances are linear (here a vacuum capacitor) voltages can be added as charges are. So the mean charge makes a mean voltage but no difference. The voltage drop comes from the charge difference.

[guggikofod]

thanks for the answers!

I didn't know that one could treat the charges in this way. Could one derive this approach from a more basic principle, like using Gauss boxes or similar? If so, I suppose this would mean that one needs to look more microscopically at what happens on each capacitor plate.

Maybe you are allowed to do this for the same reason that it is not the absolute potential which is important, but the relative potential differences?

thanks

[Enthalpy]

It's because in vacuum, Maxwell's equations are linear. That would be different with BaTiO3.