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Dielectric sphere - Find the electric field

  1. Sep 3, 2007 #1
    Consider a dielectric sphere of radius R and dielectric constant e_r. The sphere contains
    free charges that have been uniformly distributed with density rho.

    (a) Show that electric field inside the sphere is given by
    E = [rho*r /(3*e_0*e_r)]* r hat

    where r hat is the unit vector pointing in the radial direction.

    (b) Show that the electric field outside the sphere is
    E = [rho*R³]/[3*e_0*r²] * r hat

    (c) Calculate the potential V at the centre of the sphere compared to that at infinity.

    Any tips/hints that can help start me off will be appreciated.
     
  2. jcsd
  3. Sep 3, 2007 #2

    Claude Bile

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    Science Advisor

    Start with Gauss' law. Try to calculate how E varies with the radius of your spherical Gaussian surface.

    Claude.
     
  4. Sep 3, 2007 #3
    well gauss' law is just integral of E.dA = Q/e_0 since we have a sphere the SA is just 4*pi*r^2 so it gives E = Q/[4*pi*e_0*R^2].

    How does the density rho and dielectric constant e_r affect/influence this?

    Thanks everyone for helping.
     
  5. Sep 5, 2007 #4
    Use the Guass's law in presence of dielectrics.
    Within the dielectric the total charge density can be written as:
    [tex]\rho = \rho_b + \rho_f[/tex]
    where [itex]\rho_b[/tex] is the bound charge and [itex]\rho_f[/itex] is the free charge density resp.
    The Gauss's law will be modified accordingly. Read up on the electric displacement vector and you can easily find the solution.
     
  6. Sep 5, 2007 #5

    olgranpappy

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    Homework Helper

    It influences it because you don't know what Q is. As you have written it Q is the source of the electric field E, i.e. the true and total Q (both bound and otherwise). But you don't know Q because you only know the *free* charge density.

    Luckily, gauss' law will work for the electric displacement too with
    [tex]
    \int \vec D \cdot \vec dA = Q_{\textrm{free}}^{\textrm{enclosed}}
    [/tex]
     
    Last edited: Sep 5, 2007
  7. Sep 5, 2007 #6

    olgranpappy

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    Homework Helper

    ...and you also have the definition (for a linear dielectric) that
    [tex]
    \epsilon_r\epsilon_0 \vec E = \vec D
    [/tex]

    So, use what's in my first post to solve for D and then just divide to get E

    ...I might not have all the factors right since I am used to using Guassian units... but I think I got the placements of the epsilons correct...
     
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