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Capacitor AC

  1. Dec 3, 2013 #1
    1. The problem statement, all variables and given/known data
    consider a capacitor with circular plates of radius a, separated by a distance d (d<<a) and [itex]V(t)=V_{0}sin(wt)[/itex]

    a)Considering the z axis to be the capacitor axis, verify that the electric field between the plates is , in good approximation, given by [itex]\vec{E}(t)\approx E_{0} sin(wt)\hat{e}_{z}[/itex].

    What is the expression of [itex]E_{0}[/itex] ?

    In which conditions is adequate this approximation of [itex]\vec{E}(t)[/itex] ?

    2. Relevant equations

    [itex] \vec{E}= \nabla V = \hat{e}_{r} \frac{\partial V}{\partial r} + \hat{e}_{\varphi} \frac{\partial V}{\partial \varphi} + \hat{e}_{z} \frac{\partial V}{\partial z} [/itex]

    [itex]q=C V[/itex]

    3. The attempt at a solution

    I tried to find the electric field trough the above gradient formula but V has no dependance on r, phi, or z.
    I also tried trough the second equation and got [itex]\vec{E}(t)= -\frac{q_{0}}{ \epsilon_{0} A} sin(wt) \hat{e}_{z}[/itex] with [itex]E_{0}= -\frac{q_{0}}{ \epsilon_{0} A} [/itex] but then I can't explain why is it just an approximation and in which conditions is it a good aproximation
     
    Last edited: Dec 3, 2013
  2. jcsd
  3. Dec 3, 2013 #2

    berkeman

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    Staff: Mentor

    Hint -- What are the units of the electric field E? Does that give you a hint for what the peak electric field will be in terms of the voltage V and plate separation d?
     
  4. Dec 3, 2013 #3
    [itex]E_{0}=\frac{V_{0}}{d}[/itex] ?
     
  5. Dec 3, 2013 #4

    berkeman

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    Staff: Mentor

    Yes, electric field has units of Volts/meter.

    Do you have thoughts for when this will become less accurate for some capacitor configurations?
     
  6. Dec 3, 2013 #5
    hm no..
     
  7. Dec 3, 2013 #6

    berkeman

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    Staff: Mentor

    What is the traditional capacitor equation for the capacitance as a function of the Area of the capacitor and the spacing d?

    Look up that equation on Hyperphysics or Wikipedia (look under capacitor), and there should be mention of when that equation is less accurate. That will give you a clue as to when (where?) the E field equation is not so accurate...
     
  8. Dec 4, 2013 #7
    when d is not << a?

    but where is the mathematics to show it?
     
  9. Dec 4, 2013 #8

    berkeman

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    Staff: Mentor

    That is correct. When you have significant E field that is "fringing" outside of the edges of the capacitor, you can't use that simple equation for the capacitance. Instead, you need to integrate and take the fringe field into account.
     
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