Parallel plate capacitor

Homework Statement

Ok so when you have a parallel place capacitor with a charge +Q on one plate and -Q on the other, you can use Q = CV etc., My question is: what happens when the two charges differ - i.e. one plate is Q1, the other is Q2. What is the capacitance, potential difference etc?

The Attempt at a Solution

vela
Staff Emeritus
Homework Helper
I don't know offhand what the potential difference would be, but I do know that the capacitance remains unchanged. Capacitance is a function of the geometry of the conductors. It has nothing to do with the amount of charge on them.

You can see an example of this in the formula for the capacitance of the parallel-plate capacitor:

$$C = \frac{\epsilon_0 A}{d}$$

where A is the area of the plates and d is the distance between them. Those variables characterize the geometric configuration of the capacitor. Charge doesn't appear in the formula at all.

Ok thanks.
I'm trying to work out the problem on this page:

and trying to understand dadface's post about treating it as two capacitors in parallel. Doesn't Q = CV only hold for capacitors where there is +Q on one plate and -Q on the other..? I'm not sure how his derivation is meant to work...

anyone able to explain why his method works??

vela
Staff Emeritus
Homework Helper
He's modeling the sheet charge in the middle as the plates of two capacitors. Think of it as two plates, one with charge Q1 and one with charge Q2, infinitesimally separated and connected by a wire. One of these plates, say the one with charge Q1, along with one plate of the original capacitor forms a capacitor. The plate of the original capacitor has an induced charge -Q1. So you have the same magnitude of charge on each plate, as usual. Similarly, the second pair of plates have charges Q2 and -Q2.

Aha ok i see! Many thanks.

I'm trying to work it out another way, but not getting very far. any help would be much appreciated:

basically - im trying to say - the potential between the left plate and the right plate should = 0

Now let Qa be charge induced on left plate. Qb charge induced on right plate. central plate has charge q as we know.
I'm trying to work out the E field in the left hand region and the right hand region.

in the left hand region i think it should be E1 = 1/2eA (Qa + Qb - q). In the right hand region it should be E2 = 1/2eA (Qa + Qb + q)

But then when i integrate E.dl along the path from one plate to the other, i don't get the right answer... where am i going wrong?

He's modeling the sheet charge in the middle as the plates of two capacitors. Think of it as two plates, one with charge Q1 and one with charge Q2, infinitesimally separated and connected by a wire. One of these plates, say the one with charge Q1, along with one plate of the original capacitor forms a capacitor. The plate of the original capacitor has an induced charge -Q1. So you have the same magnitude of charge on each plate, as usual. Similarly, the second pair of plates have charges Q2 and -Q2.

Was just thinking about this some more. Consider the following.

Left hand plate has charge induced Qa. Right hand plate has charge induced Qb. Now Qa+Qb = 0

But you are saying we can treat the middle plate as half a plate with charge -Qa and half with charge -Qb. But that means the total charge on this plate is -Qa - Qb = 0 since we know Qa + Qb = 0, yet we know the middle plate must have total charge +q. What's gone wrong?

anyone? vela?

vela
Staff Emeritus