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Parallel plates capacitor

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Capacitor parallel plates separation $$d$$, potencial diference $$V_0$$, at the center there is a semi sphere of radius $r_0$.

    Find the potencial as function of position if $$d>>r_0$$
    2. Relevant equations
    I think that relevant equation are Gauss Law

    3. The attempt at a solution
    Mi proposed of solution is

    $$V_{plates}= \sigma /\epsilon_0 z $$

    for $$r<|r_0|$$

    Gauss Low

    $$E 4/3 \pi r^2 = \sigma 4/3 \pi r_0^2$$

    $$E= \sigma r_0^2 /\epsilon_0 r^{-2} \hat{r} $$

    $$ V_{sp} = \sigma r_0 /\epsilon_0 ( r_0/r-1) = \sigma r_0 /\epsilon_0 ( r_0/\sqrt{x^2+y^2+z^2}-1) $$

    And for $$r>|r_0|$$

    $$ V=V_{plates}+V_{sp}$$

    Help please
     

    Attached Files:

  2. jcsd
  3. Dec 7, 2015 #2

    BvU

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    Hi Alex,

    Can't really follow. It looks to me as if you assume ##\sigma## is a constant, but why would it be constant ? The plates are conducting, so the electric field has to be perpendicular to the surface at all places. I don't think you potential satisfies that.

    Note that your addition counts in a circular disk radius r0 that isn't really there !
     
  4. Dec 7, 2015 #3
    Tanks

    Some suggest for solving my problem?

    my adittion out of radius $$r_0$$ is because there is two field electrics at zone $$r>|r_0|$$, then two potential . I dont am sure of this.

    Help please
     

    Attached Files:

  5. Dec 7, 2015 #4

    BvU

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    I think I understand what you mean, and I tried to bring across that the complete field is not a simple addition obtained from two plates and half a sphere. You will have to find something that satisfies the field equations and the boundary conditions as well. Not so trivial, I grant you. Do you know about the image charge method ?
     
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