Calculating Capacitance and Electric Force in a System of Connected Capacitors

In summary, the conversation is about trying to determine the capacitance of a system with N capacitors in a cascade connection, as well as the force felt by a point charge. The suggestion is made to replace the capacitors with a single capacitance and there is a discussion about the meaning of a point charge in relation to the capacitor. The conversation then moves on to questions about the field and how to calculate it directly, as well as the role of P in the equation for D.
  • #1
shomey
30
0
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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  • #2
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?
 
  • #3
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!
 
  • #4
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...
 
  • #5
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!
 

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  • #6
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
 

FAQ: Calculating Capacitance and Electric Force in a System of Connected Capacitors

What is capacitance?

Capacitance is the ability of a system to store electric charge. It is measured in farads (F) and is represented by the letter C in equations.

How is capacitance calculated?

Capacitance is calculated by dividing the amount of stored charge (Q) by the applied voltage (V), or C = Q/V. It can also be calculated by multiplying the permittivity of the material (ε) by the area of the plates (A) and dividing by the distance between the plates (d), or C = ε * A / d.

What factors affect the capacitance of a system?

The capacitance of a system is affected by the distance between the plates, the surface area of the plates, and the type of material between the plates. It also depends on the permittivity of the material, which is a measure of how easily electric fields can pass through it.

How does capacitance impact the behavior of a circuit?

Capacitance can impact the behavior of a circuit in several ways. It can store and release energy, act as a filter for certain frequencies, and cause delays in the flow of current. It also affects the overall impedance of the circuit.

How can capacitance be increased or decreased in a system?

Capacitance can be increased by increasing the surface area of the plates, decreasing the distance between the plates, or using a material with a higher permittivity. It can be decreased by doing the opposite - decreasing surface area, increasing distance, or using a material with a lower permittivity.

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