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Bound volume and surface charges in dielectric

  1. May 6, 2014 #1
    1. The problem statement, all variables and given/known data

    2lih0g4.png

    Find surface and volume charge densities. Deduce electric field.

    2. Relevant equations



    3. The attempt at a solution

    Volume charge density:
    [tex]\epsilon_0 \epsilon_r \nabla . \vec E = \rho_f [/tex]

    Using ##\vec P = \chi \epsilon_0 \vec E = (\epsilon_r -1)\epsilon_0 \vec E##:
    [tex]\left(\frac{\epsilon_r}{\epsilon_r -1}\right) \nabla . \vec P = \rho_f[/tex]

    Thus volume charge density:
    [tex]\rho_f = \left(\frac{\epsilon_r}{\epsilon_r -1}\right) \frac{1}{r^2}\frac{\partial}{\partial r} \left[ P_0 r^3(a-r)\right][/tex]
    [tex]\rho_f = \left(\frac{\epsilon_r}{\epsilon_r -1}\right) P_0 (3a-4r)[/tex]

    Surface charge density is less tedious:
    [tex]\sigma_b = \vec P . \hat n = P_0 r(a-r)[/tex]

    Isn't the electric field within the sphere simply ##\vec E = \frac{1}{\epsilon_0 (\epsilon_r -1)} \vec P = \frac{P_0}{\epsilon_0 (\epsilon_r -1)} r(a-r) \hat r##?

    For electric field outside sphere:
    [tex]\epsilon_0 E (4\pi r^2) = \int_0^a \rho_f dr[/tex]
    [tex] E = \frac{\epsilon_r P_0}{4\pi \epsilon_0 (\epsilon_r -1)} \left(\frac{a}{r}\right)^2[/tex]
     
    Last edited: May 6, 2014
  2. jcsd
  3. May 7, 2014 #2
    Hi,

    Isn't volume charge density simply [itex]-\nabla .\vec{P}[/itex]? And since they have not mentioned [itex]\epsilon_r[/itex], you cannot use it in your solution. Moreover, they have not said that the material in linear, isotropic and homogeneous... so [itex]\vec{P}[/itex] and [itex]\vec{E}[/itex] may not be related as such.
     
  4. May 26, 2014 #3
    Ok, so the bound volume chage density is [itex]-\nabla .\vec{P}[/itex].

    The surface charge density is ##\vec P \cdot \hat n##.

    Gauss's law reads:

    [tex]\epsilon_0 \nabla \cdot \vec E = \rho_b + \rho_f [/tex]

    In this case there are no free charges, so:

    [tex]\epsilon_0 \nabla \cdot \vec E = \rho_b = -\nabla \cdot \vec P [/tex]

    Thus, ##E = \frac{1}{\epsilon_0} \vec P## ?
     
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