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Parallel electrodes (space charge density, current density)

  1. Mar 14, 2015 #1
    1. The problem statement, all variables and given/known data
    Two plane parallel electrodes are separated by a plate of thickness s whose conductivity [itex] \sigma [/itex] varies linearly from [itex] \sigma_0 [/itex] near the positive plate to [itex] \sigma_0 + a [/itex] near the negative plate.

    Calculate the space charge density [itex] \rho_f [/itex] when the current density is [itex] J_f [/itex].

    2. Relevant equations
    *see below

    3. The attempt at a solution
    I am having a tough time with this problem. Here's what I've been trying.

    the conductivity varies linearly, so setting the two plates at z=0 and z=s, [itex] \sigma = \sigma_0 + \frac{az}{s} [/itex]. I believe that gives me the correct conductivity anywhere between the plates.

    Now, I tried to relate the current density and the space charge density by trying to implement the conservation of charge equation [itex] \nabla \cdot \vec J_f = -\frac{\partial \rho_f}{\partial t} [/itex], but wasn't getting anywhere.
    I can't just sub in [itex] J_f = \sigma E = (\sigma_0 + \frac{az}{s}) E [/itex] can I? And then get an expression for the varying electric field between the plates, and plug that in. Even if I was able to modify Ohm's Law for this material, I would still have a time dependence on the space charge density term, and I know that can't be right...

    the 2nd part of the question wants you to plug in numbers (to the equation derived in this part) to calculate the space charge density. They give you [itex] \epsilon_r, J_f, s[/itex] and the two conductivities (so nothing about time).

    So now I'm more confused because [itex] \epsilon_r [/itex] is what we learned when dealing with dielectrics, but the material inserted has a conductivity...

    Does anyone have an idea how I could proceed with this problem - was I even on the right track?
     
    Last edited: Mar 14, 2015
  2. jcsd
  3. Mar 14, 2015 #2

    mfb

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    Good so far.
    You have a time-dependence in your expression. In equilibrium (!), how does the charge density depend on time?
     
  4. Mar 14, 2015 #3
    well my thinking is in equilibrium, no charges are moving so [itex] \frac{\partial \rho_f}{\partial t} = 0[/itex] so it is constant.
    But [itex]\rho_f [/itex] is what I need to solve for, so I wouldn't want to eliminate it from my equation..

    I'm going to sleep on it for now
     
  5. Mar 15, 2015 #4

    mfb

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    The charge density will re-appear elsewhere, once you have the electric field as function of the position.
     
  6. Mar 15, 2015 #5
    ok, so I couldn't see any other way on how to involve [itex]\epsilon_r [/itex] other than assuming the material is a class A dielectric...(even though it has a conductivity, which I thought was not possible). Here's what happened when I assumed this:

    Creating a gaussian cylinder from one plate up into the space density region..
    [itex] \oint \vec{D} \cdot d\vec{a} = DA = \sigma_f A + \rho_f A z [/itex]
    [itex] D = \sigma_f + \rho_f z [/itex]

    [itex] \vec{D} = \epsilon_0 \epsilon_r \vec{E} [/itex]
    [itex] \vec{E} = \frac{\sigma_f}{\epsilon_0 \epsilon_r} + \frac{\rho_f z}{\epsilon_0 \epsilon_r} [/itex]

    So if my assumptions are right, I could then plug this E (along with my varying conductivity expression) into my conservation of charge equation, giving me an expression relating [itex]J_f [/itex] and [itex]\rho_f [/itex]:
    [itex] \nabla \cdot \vec{J_f} = \nabla \cdot [(\sigma_0 + \frac{az}{s})(\frac{\sigma_f}{\epsilon_0 \epsilon_r} + \frac{\rho_f z}{\epsilon_0 \epsilon_r})] =0[/itex]

    This is the only way I could think of how to solve for the E-field (using dielectrics equations, even though I'm not sure they apply here..).
     
    Last edited: Mar 15, 2015
  7. Mar 15, 2015 #6

    mfb

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    Looks fine.

    E depends on z -> D depends on z -> between z and z+dz you need some charge -> charge density depends on derivative of D.
     
  8. Mar 15, 2015 #7
    good, thanks a lot for your help. But I'm still having an issue...
    when I solve the expression for [itex] \rho_f[/itex] I get:
    [itex]\rho_f = (\frac{1}{\sigma_0 z + \frac{az^2}{s}}) (J_f \epsilon_0 \epsilon_r - \sigma_0 \sigma_f - \frac{az^2 \sigma_f}{s}) [/itex]

    There's a couple issues with this:
    -The space charge density is infinite at z=0 (on the surface of one of the plates)
    -The book wants you to calculate [itex] \rho_f[/itex] given [itex] J_f, \sigma_0, a, s [/itex] and [itex]\epsilon_r [/itex], so they don't give [itex]\sigma_f [/itex] like I have in my expression..

    But if you think my process looks okay, I'll just ask my professor in a couple days. Maybe there's a trick or something.
     
  9. Mar 15, 2015 #8

    mfb

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    Then take a surface outside those plates, or avoid the plates completely. This looks like a pure calculation problem. Actually, your formulas look more complicated than I would expect.
    What is ##\sigma_f##?
     
  10. Mar 15, 2015 #9
    the charge density on the plates.

    Even if I ignore this charge density, i run into the same problem (division by 0 at z=0).

    The only thing I'm not confident about now is my expression for E; I've never calculated/seen an example of how to calculate E in a varying conductive material, and like i said, the only time I've used the relative permittivity was working with dielectrics, so I don't know how else to bring it into my expression
     
  11. Mar 15, 2015 #10

    TSny

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    Why can't you solve this for E to get E as a function of z? (Maybe I'm missing something.)

    Once you have E(z), you can calculate ∇⋅E.
     
  12. Mar 15, 2015 #11
    wow, yeah that makes sense to me!

    but why would they give me the relative permittivity [itex] \epsilon_r [/itex]?
    Since [itex] \nabla \cdot \vec{E}= \frac{\rho}{\epsilon_0}[/itex] , am I just supposed to add the relative permittivity to this equation? Is that even legal?
     
  13. Mar 15, 2015 #12
    yeah I think it's okay to do that
     
  14. Mar 15, 2015 #13

    TSny

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    ##\vec{\nabla} \cdot \vec{E} = \frac{\rho_f + \rho_b}{\varepsilon_0} = \frac{\rho_f}{\varepsilon_r \varepsilon_0}## for a linear dielectric where ##\varepsilon_r## does not depend on position. (##\rho_b = ##bound charge density.)
     
  15. Mar 15, 2015 #14
    yep! thanks TSny. I guess (non perfect) dielectrics are allowed to have a conductivity
     
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