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Electromagnetics - Parallel Plate Capacitor [so stuck ]

  1. Sep 30, 2006 #1
    I'm stuck.

    Here's the question,

    Q: The upper and lower conducting plates of a large parallel-plate capacitor are separated by a distance [itex] d [/itex] and maintained at potentials [itex] V_0 [/itex] and [itex] 0 [/itex], respectively. A dielectric slab of dielectric constant [itex] 6.0 [/itex] and uniform thickness [itex] 0.8d [/itex] is placed over the lower plate. Assuming neglible fringing effect, determine
    a) the potential and electric field distribution in the dielectric slab,
    b) the potential and electric field distribution in the air space between the dielectric slab and the upper plate,
    c) the surface charge densities and the upper and lower plates

    A (what I have so far):

    The governing electrostatic equation is:
    [tex] \nabla^2 = \frac{-\rho}{\epsilon} [/tex]

    Between the slabs [itex] \rho = 0 [/itex],
    [tex] \nabla^2 V = 0 [/tex]

    So if we first examine the potential:

    [tex] \nabla^2 V = \frac{d^2}{dz^2} = 0 [/tex]
    Solving this ODE for the two regions of interest yields:

    [tex] V(z) = C_1 z + C_2 \,\,\,\, 0 \leq z < 0.8d [/tex]
    [tex] V(z) = C_3 z + C_4 \,\,\,\, 0.8d \leq z \leq d [/tex]

    (confusion 1)
    I'm not sure about the domain for the potential function. How do I describe the potential at [itex] z = 0.8d [/itex]?
    Does [itex] V(0.8d) = C_1 z + C_2 [/itex]
    or
    [itex] V(0.8d) = C_3 z + C_4 [/itex]

    and why? (or a hint is fine)

    So, since we have C1, C2, C3, and C4 we are going to need two more equations, and four boundary conditions. I'm confused about the boundary conditions, and I think if I get a hint here, I'll be able to come up with the other two equations.

    So looking at the boundary conditions, (this is definitely where I am confused (part two))

    The obvious:
    (1) [tex] V(x,y,z=0) = 0 [/tex]
    (2) [tex] V(x,y,z=d) = V_0 [/tex]

    Now what about (3) and (4)?
    There needs to be something at the media crossing right? So is this correct?

    [tex] \hat n_2 \cdot (\vec D_1 - \vec D_2) = \rho_s [/tex]
    [tex] \rho = 0 [/tex]
    [tex] \hat n_2 = -\hat z [/tex]

    Thus,
    (3) [tex] D_{1z} = D_{2z} [/tex]

    any help on the above bold spots would be amazing. Thank you!
     
    Last edited: Sep 30, 2006
  2. jcsd
  3. Oct 1, 2006 #2

    Astronuc

    User Avatar

    Staff: Mentor

    One needs a continuity boundary condition at 0.8d, e.g. V(0.8d). What about continuity of [itex]\vec{E}[/itex] or [itex]\vec{D}[/itex]?

    If one has 4 constants, then one needs four linearly independent equations with which to solve for those 4 constants.
     
  4. Oct 1, 2006 #3
    Thanks for the reply Astronuc. So I've been digesting what you have said, and I still can't wrap my fingers around the V(0.8d) part.

    I believe I understand the boundary conditions at the interface better now, and will state what I have believed to learn during this time.

    So the two boundary conditions at the interface are:
    [tex] E_{1t} = E_{2t} [/tex]
    [tex] D_{1n} = D_{2n} [/tex]

    So since I have an expression for the voltage, I find the e-field, and then the electric displacement. So my third boundary condition will be:

    [tex] D_{1n}(0.8d) = D_{2n}(0.8d) [/tex]

    I do not understand how to arrive at the voltage boundary condition.
    My best guess is,
    V1(0.8d) = V2(0.8d), where V1 denotes the voltage expression that has the domain restricted to the dielectric slab, and V2 denotes the voltage expression that has the domain restricted to the air space between the dielectric and the conductor.

    But, I am unsure...
     
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