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Homework Help: Electric Field in dielectric

  1. May 5, 2014 #1
    1. The problem statement, all variables and given/known data


    Find the electric field inside dielectric and surface charge density on plates.

    2. Relevant equations

    3. The attempt at a solution

    [tex]\rho_b = -\vec {\nabla} . \vec{P} = -\vec {\nabla} . \epsilon_0 \chi \vec E = -\vec {\nabla} . (\epsilon_r - 1)\epsilon_0 \vec E[/tex]

    [tex]\int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec {\nabla} . \vec E dV[/tex]

    [tex] \int \rho_b dV = (1-\epsilon_r)\epsilon_0 \int \vec E . d\vec S [/tex]

    [tex]\rho_0 \int (d^2 - x^2) A dx = (1-\epsilon_r)\epsilon_0 EA[/tex]

    [tex]E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0} \int_{x'}^{d} d^2-x'^2 dx'[/tex]

    [tex]E = \frac{\rho_0}{(1-\epsilon_r)\epsilon_0}\left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right][/tex]

    To find surface charge density,

    [tex]\epsilon_0 \nabla . E = \rho_f + \rho_b[/tex]
    [tex]\epsilon_0 \int \nabla . E dV = \int \rho_f dV + \int \rho_b dV[/tex]
    [tex]\epsilon_0 \int E.dS = \sigma A + \int \rho_b A dx[/tex]
    [tex]\epsilon_0 EA = \sigma + \int \rho_b dx[/tex]


    [tex]\sigma = \epsilon_0E - \epsilon_0(1-\epsilon_r)E = \epsilon_0 \epsilon_r E[/tex]
    [tex]\sigma = \frac{\rho_0 \epsilon_r}{1-\epsilon_r} \left[ \frac{2}{3}d^3 - d^2x + \frac{1}{3}x^3\right][/tex]
  2. jcsd
  3. May 6, 2014 #2


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    The embedded charge density given in the problems is "free charge" density ##\rho_f## of the system, not bound charge density ##\rho_b##.

    In your integrals, you are integrating from ##x'## to ##d##. But the upper limit can be tricky to deal with since there is bound surface charge density on the surface of the dielectric at ##x = d## as well as free surface charge density on the surface of the conductor at ##x = d##.

    There is quite a bit of symmetry in the problem. Can you use the symmetry to tell you anything about D and E at x = 0? If so, you might want to set up your integrals from ##x = 0## to ##x = x'## and avoid ##x = d##.
  4. May 6, 2014 #3
    Ah, I see the problem now.

    Gauss's law in Dielectric reads:
    [tex]\nabla . D = \epsilon_0 \epsilon_r \nabla . E = \rho_f[/tex]

    [tex]\epsilon_0 \epsilon_r EA = \rho_0 \int_0^{x'} (d^2 - x'^2) A dx' [/tex]

    [tex]E = \frac{\rho_0}{\epsilon_0 \epsilon_r}\left[d^2x - \frac{1}{3}x^3\right][/tex]

    For the surface charge, we must compute the bound surface charges on the dielectric first.


    Gauss's Law reads:

    [tex]\epsilon_0 EA = Q - Q_b[/tex]
    [tex]\epsilon_0 EA = \sigma A - A\vec P . \hat n[/tex]
    [tex]\epsilon_0 E = \sigma - \epsilon_0 (\epsilon_r -1)E[/tex]
    [tex]\sigma = \epsilon_0 \epsilon_r E[/tex]

    Or in other words, ##E = \frac{\sigma}{\epsilon_0\epsilon_r}##, the electric field is reduced by a factor of ##\epsilon_r## due to the presence of the dielectric's polarized field.

    Electric field at d is ##\frac{2}{3}\frac{\rho_0}{\epsilon_0\epsilon_r} d^3##.
    Thus, ##\sigma = \frac{2}{3}\rho_0 d^3##.
    Last edited: May 6, 2014
  5. May 6, 2014 #4


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    I get essentially the same result except for some signs. I get a positive amount of bound surface charge at the surface of the dielectric (due to the polarization of the medium). I get a negative surface charge density on the surface of the metal plate.

    If I pick a Gaussian pill box as you did, I would write Gauss' law as

    ##-\epsilon_0 E_x A = Q_{tot} = Q + Q_b = \sigma A + P_x A = \sigma A +\epsilon_0(\epsilon_r - 1)E_x##
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