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Homework Help: Maximum voltage in a capacitor with 2 dielectrics

  1. Jul 25, 2010 #1
    1. The problem statement, all variables and given/known data
    We have two infinite conductor plates, with two dielectrics (with permitivities K1 and K2) put together between them so that the capacitor may be taken as two capacitors connected in series. We know that the maximum electric field for the first dielectric is Er1, and we also know the thicknesses of both dielectric layers (we'll call the thicknesses d1 and d2).

    A) What's the value of the maximum electric field for the second dielectric, so that both will break at the same time?
    B) Which voltage do we need to apply in order to get such field?

    (When I say "maximum electric field", I'm referring to the electric field required to break the dielectric and turn it into a conductor)

    2. Relevant equations
    Let E0 be the electrical field applied to the capacitor, and Vr be the required voltage to break both dielectrics.

    [tex]{E_r} = \frac{{{E_0}}}{{{k_i}}}[/tex] (with i=1,2)
    [tex]V = Ed[/tex]

    3. The attempt at a solution

    [tex]{E_0} = {E_{r1}}{k_1}[/tex]


    [tex]{E_0} = {E_{r2}}{k_2}[/tex]

    We can write

    [tex]{E_{r1}}{k_1} = {E_{r2}}{k_2}[/tex]


    [tex]\frac{{{E_{r1}}{k_1}}}{{{k_2}}} = {E_{r2}}[/tex]

    which gives us the required field to break the second dielectric. I assume that if we apply an electric field of at least E0 we'll break them both. Which leads us to calculate the required voltage:

    [tex]{V_r} = {V_1} + {V_2}[/tex] (consider both dielectrics as two capacitors connected in series)

    threrefore, the required voltage would be

    [tex]{V_r} = {E_{r1}}{d_1} + {E_{r2}}{d_2}[/tex]

    Is my procedure correct? Thanks a lot.
  2. jcsd
  3. Jul 25, 2010 #2
  4. Jul 26, 2010 #3


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    Homework Helper

    We do not apply electric field. We apply voltage. Your E0 is the "electric displacement" which is equal to the surface charge density. As the charge is the same on the series capacitors you can say that E0 = Er1K1=Er2K2.

    Anyway, the results of your solution are correct.

  5. Jul 26, 2010 #4
    Thanks a lot.

    Just out of curiosity, how would the answer differ if the dielectrics were in parallel? I mean besides the voltage being the same for both instead of the sum.
  6. Jul 26, 2010 #5


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    If the dielectrics were in parallel, the capacitors were in parallel, too, and both had the same thickness. So the electric field intensity were the same E=V/d for both. That part would brake down where V/d is greater than the maximum allowed electric field.
    The voltage were the same, but the surface charge density were different , being equal to Ki*V/d.

  7. Jul 26, 2010 #6
    I didn't understand the last part. Does the Ki*V/d come from P=[tex]\epsilon[/tex]0*Xe*E, which is equal to the surface density? I assume you're calling =[tex]\epsilon[/tex]0*Xe "Ki", but in that case wouldn't it be Ki*V/(d*Ke), Ke being the dielectric's permitivity?
  8. Jul 26, 2010 #7


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    I denoted the absolute permittivity of the i-th dielectric by Ki, as you used this notation for the permittivity in your first post. You might have meant relative permittivity. I would note the (absolute) permittivity of a dielectric by ε. If the surface charge density on a metal is σ, the electric field intensity in the dielectric at the neighbourhood of the metal surface is E=σ/ε.
    E=V/d inside both capacitors. As V is the same for the parallel capacitors and d is supposed to be the same, too, the electric field intensity is also equal. But this means that the surface charge density is different on the plates of the capacitors filled with different dielectric.

  9. Jul 26, 2010 #8
    Oh, I see! You meant the surface density for the free charges in the plates, not the induced density in the dielectrics.

    Thanks a lot!
  10. Jul 26, 2010 #9


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    Yes, of course. The charge of the capacitor is on the plates.

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