What would be the capacitance of a particle on a plate?

[Javier Lopez]

Homework Statement

There is a charged particle at height h over a circular area of radii R, then I have to calculate the capacitance. This is useful when somebody want to calculate energy harvesting, antennas, sensors that measures ions in plasma devices and particle accelerators.
https://lh5.googleusercontent.com/_-xHmEdtx4XEzINcGpKFP-fj3_32U7U4xYBm4pFLUqTMQlHc_zC_n-WsELCXBnDHBY-YqKgLecMmePFSRpde=w1920-h950 [/B]

Homework Equations

Q=CV
Gauss law, integral form that can be seen in (1)

The Attempt at a Solution

https://lh5.googleusercontent.com/VcIzJXKDsaXa8zi5Q2DS-R9z1DkDsRNvtKsq6lFQGvfzdnjcT04kVL9xhNk2Vi4-HZfd4I-YWiFM1hU2D7QK=w1920-h950

I am not happy with my solution because the plate down has also charges and I do not know if those charges have to be included in the equation
Also there is the assumption that all electric field lines goes straight to the plate that could be true if the particle is close to it.

 

[BvU]

https://lh5.googleusercontent.com/VcIzJXKDsaXa8zi5Q2DS-R9z1DkDsRNvtKsq6lFQGvfzdnjcT04kVL9xhNk2Vi4-HZfd4I-YWiFM1hU2D7QK=w1920-h950 is all I see. Twice.

Anyway: what is the definition of capacitance for your configuration ?

And: Hello Javier,

 

[mfb]

You can't link to googleusercontent.com like that. You might see the images, but no one else has access to them.

You can use the "upload" button here to add images to your post.

 

[Javier Lopez]

I hope you can see now, regards!

 

[BvU]

Yes I can see it now, but I don't understand one iota. What is it you are calculating ? One moment you have $E(x)$, so $V$ depends on $x$, and a little later you take $V$ in front of the integral -- as if it does not depend on $x$ ?

What is the charge stored on the disc ?
Could you complete the problem statement: is the disc conducting ? Is it grounded ? What is the diameter of the charged particle Q ?

Did you compare your result with that for 'Sphere in front of wall' ?

(Maxwell, J. C. (1873). A Treatise on Electricity and Magnetism. Dover. pp. 266ff.)

 

[Javier Lopez]

I used that the voltage is constant in the plate as long as it is conductive: you can use a multimeter to realize that always the voltage between a metal part is always 0 (unless a high power pulse is applied), so not depends on x
Then I stated that V between particle and plate is constant.
As long as at far distance the charge is 0, then I stated that Q induces at the plate a charge -Q that will have uniform density (if not uniform the voltage between plate parts is not 0).

I set the equation at wikipedia in an excel table and I obtained that capacitance is about 2*pi*epsilon*a where a is de radius of the charge that is almost 0. Then my equation should be wrong.

I look at the Maxwell print and not found that, I found the capacitance of the two spheres.

 

[BvU]

Maybe you want to look at the method of images
(or here)

 

[Javier Lopez]

Thank you after looking at that method, the problem is I took in account only the field due the charged particle and not the field coming from the plane, so fortunately my formula is wrong
(if not a big problem could happen because the capacitance of a zero sized charged would not be 0)
I have a new thread: to use the images method and write here the result, unless somebody do it before!

Could help if the webmaster added a formula app compatible with the microsoft or latex one

 

[Javier Lopez]

I was stuck until BuV recommended me to use the Image Method, now it is easy to be solved:

 

[BvU]

Some notes:

The image method is only valid in that part of space where the real charges are sitting: the +Q and the surface charge on the plate. Not in the volume for which $x>a$.

From symmetry you have $V = 0$ at $x=0$. This is the boundary condition that is exploited by this image charge method.

Is it clear to you that the potential at a point charge is infinite ? All you do in your calculation is re-discover that.

The 'Sphere in front of wall' case in link 5 should be a lot more interesting (but also a lot more complicated) in the sense that it leads to a capacitance -- probably not in relation to the subjects you mentioned in your post #1; for those the induced charge distribution on the plate may be more interesting.

 

[Javier Lopez]

Ok, integral limits should be between x=0 and x=h not 2h, but result is ok. I just reinvented the wheel but enjoyed :)

 

[Javier Lopez]

If somebody want to try to calculate the charge density in the plate, is free to try, I will wait to volunteers!

 

[BvU]

If somebody want to try to calculate the charge density in the plate, is free to try, I will wait to volunteers!

Instructive ! 2nd Link in post #7 (here) pages 9 ...12 (for a big plate)