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Finding the bound charge in a dielectric ATTEMPT 2

  1. Nov 17, 2013 #1
    1. The problem statement, all variables and given/known data
    The space between the plates of a parallel plate capacitor is filled with a dielectric material whose dielectric constant ϵr varies linearly from 1 at the bottom plate (x=0) to 2 at the top plate (x=d). The capacitor is connected to a battery of voltage V. Find all the bound charge, and check that the total is zero.


    2. Relevant equations

    [itex]C=\frac{Q}{V}[/itex]

    [itex]C=\frac{Aε_{0}}{d}[/itex]

    [itex]D=ϵE[/itex]

    [itex]D=ε_{0}E+P[/itex]

    [itex]\int D\bullet da = Q_{f}[/itex]

    [itex]σ_{b}=P⋅\widehat{n}[/itex]

    [itex]ρ_{b}=−∇⋅P[/itex]



    3. The attempt at a solution

    First I found [itex]ε_{r}[/itex] as a function of x, [itex]ε_{r}=\frac{x}{d}+1[/itex].
    Assuming that the area of each plate is A, I said that the bound charge must be [itex]Q_b=A(\sigma_b+\int\rho_{b}dx)[/itex].
    Next, using [itex]D=ε_{0}E+P[/itex] and [itex]D=εE[/itex] I get that [itex]P=D(\frac{x}{x+d})[/itex].
    Using [itex]\int D\bullet da = Q_{f}[/itex] I get that [itex]D=\frac{Q_{f}}{A}[/itex].
    Nowing using [itex]σ_{b}=P⋅\widehat{n}[/itex] and [itex]ρ_{b}=−∇⋅P[/itex] I get that [itex]\sigma_b=0[/itex] at x=0, [itex]\sigma_b=-\frac{Q_f}{2A}[/itex] at x=d and [itex]\rho_b=\frac{Q_f}{A}\frac{x}{(x+d)^2}[/itex].
    I integrate [itex]\rho_b[/itex] from 0 to d and I get [itex] \frac{Q_f}{A}(\ln 2 -\frac{1}{2})[/itex].
    So, now plugging those values into [itex]Q_b=A(\sigma_b+\int\rho_{b}dx)[/itex] I get [itex]Q_f(\ln 2 -1)[/itex], which only equals zero when there is no free charge on the plates of the capacitor, so I know I am doing something wrong. I am pretty sure I am going about this completely wrong, but I don't see any other way to do it. Any help would be really appreciated. Thank you.

    Also, using [itex]C=\frac{Q}{V}[/itex], [itex]C=\frac{Aε_{0}}{d}[/itex], and [itex]D=ϵE[/itex] I can get the capacitance [itex]C=\frac{A\epsilon_0}{d\ln 2}[/itex] and [itex]Q_f=\frac{A\epsilon_{0}V}{d\ln 2}[/itex] but I don't see how that will help me find the bound charge.
     
    Last edited: Nov 17, 2013
  2. jcsd
  3. Nov 17, 2013 #2

    TSny

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    I think that's correct.

    Should ##\sigma_b## have a negative sign at x = d? I guess it depends on your sign conventions and which plate is positively charged.

    I don't agree with the numerator ##x## in this expression.
     
  4. Nov 17, 2013 #3
    I get negative because [itex]\hat{n}=-\hat{i}[/itex] since x is increasing as you move from the first plate (x=0) to the second (x=d).

    I just re-did my calculation for [itex]\rho_b[/itex] and I got d in the numerator now, and the whole thing has a negative sign. This will change my [itex]\int \rho_b dx[/itex] and hopefully give me an answer that makes sense. Is that what you got?

    Thanks.
     
  5. Nov 17, 2013 #4

    TSny

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    Yes, I got ##d## in the numerator instead of ##x##. I got positive for ##\sigma_b## at ##x = d## and I got ##\rho_b## to be negative. I assumed the lower plate (x = 0) is the positively charged plate so that D, E, and P all point in the positive x direction.
     
  6. Nov 17, 2013 #5
    Great, I am getting that [itex]Q_b=0[/itex] now, with the bound volume charge equal to negative one half and the bound surface charge to be negative one half. Thank you so much.
     
  7. Nov 17, 2013 #6

    TSny

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    OK, but how do two negative quantities add to zero?
     
  8. Nov 17, 2013 #7
    Sorry, that was a typo. Only the volume bound charge was meant to be negative.
     
  9. Nov 17, 2013 #8
    And I meant one half [itex]Q_f[/itex] for both cases.
     
  10. Nov 17, 2013 #9
    And I realized my mistake with the negative sign on [itex]\sigma_b[/itex], I was thinking of the surface of the plate rather than the surface of the dielectric.
     
  11. Nov 17, 2013 #10

    TSny

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    Great. Good work.
     
  12. Nov 17, 2013 #11
    Thanks again.
     
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