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Electric field in leaky capacitor model

  1. Aug 7, 2017 #1
    Hi everyone.
    I need to calculate E-field(AC source / DC/ point charges) in leaky capacitor( epsilonr= a dielectric with relative permittivity and finite conductivity). But I do not how to calculate or write the equations for them. any idea or reference is very welcome.
     
  2. jcsd
  3. Aug 9, 2017 #2
    If the capacitor is not connected to a power supply, then the leakage means that it will be almost empty.
    If it is connected to a power supply, the voltage on the capacitor is equal to that of the supply.
    The strength of the electric field is equal to the voltage divided by the spacing, in Volts per metre.
     
  4. Aug 9, 2017 #3
    Thank you for your response.
    The leaky capacitor is connected to the current source (AC) via two metal electrodes. But How does the conductivity and relative permitivvity play the roles in this scenario?
     
  5. Aug 9, 2017 #4

    CWatters

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    The relative permittivity effects the capacitance...

    C = kε0A/d

    Perhaps you can model the leakage as a resistor in parallel with the capacitor?
     
  6. Aug 10, 2017 #5
    Thank you for your response.
    The problem is I got confused to use the equations...to include the resistor( in parallel ) in electric field equation...
    I came to the idea to use integral of the electric current ( AC , sine ) over a time duration ( like t1 to t2) which can give me the Q , this current source is connected to the leaky capacitor, then assume this Q is point charge. Then I can calculate the E-field of this point charge in the capacitor.
    But the problem is:
    If I assume this is a perfect dielectric (capacitor): E=Q/(4*Pi*epsilon0*epsilonr).... is not that true?
    In this case how can I consider the ( conductivity) of the capacitor in this equation...I need a leaky capacitor model ..which equation and formula would work?????????.
     
  7. Aug 10, 2017 #6

    berkeman

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    Staff: Mentor

    The "leakiness" of the capacitor does not affect the electric field, unless your AC source has a moderate source impedance. That would lower the voltage across the cap, which would lower the electric field.

    As mentioned previously, the E-field in a capacitor is just the voltage divided by the spacing (at least when you are inside the capacitor away from the edges). Can you say more about what you are trying to do and why?
    Looks like you have a stuck keyboard key... :wink:
     
    Last edited: Aug 10, 2017
  8. Aug 10, 2017 #7
    I think my keyboard does have problem. Since, I cannot edit the post. Please accept my apology.
    Thanks
     
  9. Aug 10, 2017 #8

    berkeman

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    Staff: Mentor

  10. Aug 10, 2017 #9
    Thank you for your response.
    in this case if assume :
    the leaky capacitor is a circle and on two different pints (two electrodes) .with leaky capacitor it seems to me the charges can flow through the capacitor. this makes me confused.
    (I=sin(w1*t))
    1)Can you please let me know if you want to write the equations( respect to x-y directions). what would you write?
    2)would you include conductivity and relative permittivity together? and how to write the electric field equations(say which formula)?
    thanks
     
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