What is the effect of unequal charges on a parallel plate capacitor?

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Homework Help Overview

The discussion revolves around the behavior of a parallel plate capacitor when the charges on the plates are unequal, specifically examining the implications for capacitance and potential difference. The original poster questions how to approach the problem when one plate has charge Q1 and the other has charge Q2, contrasting this with the standard case of equal and opposite charges.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the idea that capacitance is determined by the geometry of the capacitor rather than the charge on the plates. There is discussion about modeling the system as two capacitors in parallel and questioning the validity of applying the formula Q = CV in this context. Some participants express confusion about the derivation and the implications of charge distribution on the plates.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of capacitance and charge distribution. Some have offered interpretations of the problem, while others are seeking further clarification on the underlying concepts and assumptions. There is a recognition of the complexity of the situation, particularly regarding the charge interactions and the potential difference.

Contextual Notes

Participants note the potential confusion arising from the original problem's formulation and the assumptions about charge separation and grounding conditions. There is an acknowledgment of the need for clearer definitions and constraints to facilitate understanding.

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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?


Homework Equations





The Attempt at a Solution

 
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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:

[tex]C = \frac{\epsilon_0 A}{d}[/tex]

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:

https://www.physicsforums.com/showthread.php?t=426237

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??
 
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 - I am 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?
 
vela said:
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?
 
Honestly, I'm not very interested in the original problem because I don't think it was well posed, precisely because of the questions you raised, so I think you should post your thoughts and questions in the original thread on that problem. In that thread, ehild touched on your observation when he talked about whether the plates were grounded or not.
 
  • #10
ok but how is one to solve it...? Assuming it is a thick enough plate for charge to separate as you say,,>
 

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