Calculating Capacitance and Electric Force in a System of Connected Capacitors

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Homework Help Overview

The discussion revolves around calculating the capacitance of a system of connected capacitors and understanding the electric force experienced by a point charge in this context. Participants are exploring the configuration of capacitors, particularly in series, and the implications of treating capacitor plates as point charges.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are considering the series connection of capacitors and questioning the nature of the charge on the capacitor plates. There is discussion about the possibility of a capacitor lacking traditional metal plates and the implications of treating the upper plate as a point charge for force calculations.

Discussion Status

Some participants have provided clarifications regarding the series connection of capacitors and the interpretation of charge distribution. Others are exploring deeper questions about electric fields and the conditions under which certain assumptions hold true, indicating a productive exchange of ideas without a clear consensus.

Contextual Notes

There are references to specific assumptions about the capacitor's structure and the nature of electric fields, with some participants expressing uncertainty about the definitions and properties involved.

shomey
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I'm trying to figure out the capacitance of the attached system.

may I look at this problem as N capacitors attached in a cascade connenction?

I'm also trying to figure out the force that will be felt by the point charge.
I thought about using the image method... could you thing of a better solution? I'm worried it will be very long and frustrating...

thanks!
 

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Yes, all those are in series. You can replace them by a single capacitance.

I don't understand the statement "..and can be seen as a point charge Q". Do you mean to say that the charge on the capacitor plate is Q?
 
maverick280857 said:
Yes, all those are in series. You can replace them by a single capacitance.

I don't understand the statement "..and can be seen as a point charge Q". Do you mean to say that the charge on the capacitor plate is Q?

Thanks very much for the help!

sorry for the bad explanation.
what i was told is that the electrode of the capacitor is a point charge Q.
I guess it means two things:
1. the charge on the capacitor's upper plate is Q.
2. the capacitor's upper plate is very small on the horizontal axises - which sounds very strange...

This leads me to my next question:

could there be a capacitor without two metal plates on its two ends?
It is not drawn on the picture I got (only N dielectric plates), but maybe I was supposed to assume that they are there...

Again - thanks very much for the help!
 
shomey said:
Thanks very much for the help!

sorry for the bad explanation.
what i was told is that the electrode of the capacitor is a point charge Q.
I guess it means two things:
1. the charge on the capacitor's upper plate is Q.
2. the capacitor's upper plate is very small on the horizontal axises - which sounds very strange...

This leads me to my next question:

could there be a capacitor without two metal plates on its two ends?
It is not drawn on the picture I got (only N dielectric plates), but maybe I was supposed to assume that they are there...

Again - thanks very much for the help!

Thought about it a little more and I guess what they've ment in the question is that the upper electrode could be considered a point charge when calculating the force it feels...
 
Hey guys!

I think I've solved it, attached is my solution.

I have some last little points I would love to understand better:

1) why is the field D contant though the capacitor? (I've read it in a notebook but could not understand why this is true - they've claimed it is because of the symmetry of the problem...)
2) In the solution I've calculated E_n using Gauss envelopes, and then saw that D is constant, how can I calculate D directly?
3) sometimes we write D=eps*E+P and sometimes we ommit P... why is that? how could I explain this?

thank you very much for the help!
 

Attachments

Re 1&2. You have, for all practical purposes done a direct computation.Otherwise, look at the integral for the field, and you'll be direct, and 2, you'll see the important symmetry in action

Regards,
Reilly Atkinson
 

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