Parallel Plates Different areas

In summary, the person is trying to solve for the capacitance between two plates of different area. They say that if one plate is smaller than the other or infinite, then the calculation would become easier. However, they still need to solve for the e-fields first. Once they have the e-fields, they can solve for the capacitance.
  • #1
junglepeanut
12
0
Hi
I am trying to solve this problem. I would like to be able to solve for the capacitance between two parralel plates of different areas.

First I tried thinking of them as concentric cylindrical shells but twisting them to do this is not the same thing I realized because the distance between the two plates would be huge, where if they are left straight it is a very small differnce

So I am thinking of doing something similar to say Griffiths problem 3.9, but with plates instead of line charge. And i am stuck because I can tell already this is going to be hard math. If I assume one plate is really small compared to the other or if one is say infintie will this help solving the problem?

Any help would be appreciated, I think if I could figure the E field out for this case I could then find the capacitance without much hassle.

One plane is E=(surface charge)/(2*epsilon not) ...infinite
the other should be E=(surface charge)(ab)/(4*pi*epsilon not*r^2), where ab is the area of the plate
if the top plate is not infintie then I could use the bottom formula twice but with a2b2 right?

Then To find the efield in between the two plates I would just have to use superpostion to add the fields right?

Integrate then from bottom plate to top plate for voltage, and have q divided by that, this should be my capacitence ehh?


Thanks
 
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  • #2
I doubt that a problem involving a capacitor with two different area plates would be in any introductory E&M class. You wouldn't have the luxury of assuming a uniform field in between, even if the plates are close together, since that would seem to defeat the whole purpose of the cap plates having two different areas in the first place. You would need to take into account edge effects and such, and the calculation would become very involved. I think I've worked a similar problem, but it took around six or seven pages of detailed calculations that I can't really show you in a one paragraph post, but I could give you the general gist of it if you are so inclined.
 
  • #3
I would really appreciate any help you could give. It is not for an introductory E&M class, (I just finished my junior year of e&M not with flying colors either.) I was asked to find the answer, and I did not need to work it either just get the answer, but I am interested in solving the problem.

I don't mind simplifying it some by say assuming one plate is a line and the other is infinite plane, or say that one plate is much smaller than the other. I am interested in solving the actual problem i.e. the capacitance of two diffrent area plates.

I would say sovle for the potential, to get that I need the e-fields first.

I am pretty sure my efield calcs above overly simplified it or are just plain wrong.

If you don't mind giving me a hint to start off I would appreciate it.

Then once I get the e-field, Am I going to have to use my boundary conditions because of the fact that the fields will not cancel each other out above and below like ina parrallel plate capacitor. AHH let's not get ahead of myslef.

Just a hint to get started please. Then once I have solved your hint if you don't mind I will post it and try working the next step.
 
  • #4
I don't remember many specific details of how it was done, but to get you started, a way I could think of to avoid the potential calculation altogether (you will unfortunately still need the E field however) would be to use the energy density of an electric field

[tex] \mu_e = \frac{1}{2} \epsilon_0 E^2 [/tex]


Integrate over all space(my guess it involves a triple integral or something since the E field will not be uniform); this integral will give you the total potential energy of the capacitor's configuration. Then simply set this to

[tex] U_E = \frac{Q^2}{2C} [/tex]

which relates the potential energy to the charge on one plate of your capacitor, and the capacitance itslef. Do some algebra and solve for C. This problem is beginning to interest me now, figure out a specific geometry for the plates, and we can find an E field from there.
 
  • #5
Sorry I am really busy with work I will get back to you hopefully tonight, by the way just a little info I am normally in Pomona,CA but for a little while I will be near San Jose so when were done maybe I can get you lunch or something. K

Have fun thinking ab and cd for the areas of the two respective plates, don't know if integrating over all areas will be applicable because at areas of the plates we will get infiniti right? Maybe not but I think so, I think we need to treat as method of images to find voltage maybe? Maybe I will just do a line charge to refresh and check my answer before moving on to this either way we need to find the e-field right, I don't know after work I will check.
 
  • #6
Just thinking about it topically, (sorry I haven't had a chance to work on it much, finals ) I think the best you can do is calculate a capacitance as a function of an x,y position on the plane of one of the plates. Or just an average capacitance; since the capacitance varies over position on the place. Is this what you are looking for, or is it a single capacitance of some sort?
 
  • #7
I am really looking to be able to find the capacitance of the whole, somehting I can measure after creating it.

Lets say that plate A has area ab. And plate B has area ef ,
Lets let d be the distance between the plates.

I am going to get right back to this later tonight first I want to check with some of the grad students around me see if they maybe have a shortcut I really hope respond later tonight, don't worry I am workoing on this slowly as it is summer break and work has got me busy seven days a week.
 
  • #8
Also sorry if some of my posts arent so intelligible I tend to think way faster than I can type and then I think i have typed something but I haven't typed it at all, I know I can go back and edit but 90% percent of the time i realize I was rambling anyways so why bother fixing it.
 

Related to Parallel Plates Different areas

1. What is the purpose of using parallel plates with different areas in an experiment?

The use of parallel plates with different areas allows for the creation of an electric field with a varying strength. This can be useful in experiments that require the manipulation of electric fields, such as in studies of electrostatic forces and capacitance.

2. How do the areas of the parallel plates affect the electric field between them?

The strength of an electric field is directly proportional to the area of the parallel plates. This means that larger plates will produce a stronger electric field compared to smaller plates, given they have the same charge and distance between them.

3. Can parallel plates with different areas be used to measure the strength of an electric field?

Yes, parallel plates with different areas can be used to measure the strength of an electric field. By using a known voltage and measuring the capacitance, the electric field strength can be calculated using the equation E = V/d, where E is the electric field strength, V is the voltage, and d is the distance between the plates.

4. How does the distance between the parallel plates affect the electric field?

The strength of the electric field is inversely proportional to the distance between the parallel plates. This means that as the distance between the plates increases, the electric field strength decreases.

5. Can parallel plates with different areas be used to demonstrate the concept of capacitance?

Yes, parallel plates with different areas can be used to demonstrate the concept of capacitance. A larger area between the plates results in a higher capacitance, as there is a larger surface area for the electric charge to accumulate. This can be seen in the equation C = εA/d, where C is capacitance, ε is the permittivity of the material between the plates, A is the area, and d is the distance between the plates.

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