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Parallel Plates Different areas

  1. Jun 17, 2004 #1
    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 gonna 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 with out 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?

  2. jcsd
  3. Jun 18, 2004 #2


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    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.
  4. Jun 19, 2004 #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 dont 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 dont 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 feilds will not cancel each other out above and below like ina parrallel plate capacitor. AHH lets not get ahead of myslef.

    Just a hint to get started please. Then once I have solved your hint if you dont mind I will post it and try working the next step.
  5. Jun 20, 2004 #4


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    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.
  6. Jun 23, 2004 #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, dont 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-feild right, I dont know after work I will check.
  7. Jun 26, 2004 #6


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    Just thinking about it topically, (sorry I haven't had a chance to work on it much, finals :yuck: ) 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?
  8. Jun 26, 2004 #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, dont worry I am workoing on this slowly as it is summer break and work has got me busy seven days a week.
  9. Jun 26, 2004 #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 havent 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.
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