1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Potential of sphere in electric field

  1. May 29, 2014 #1
    1. The problem statement, all variables and given/known data


    Part (a): Find potential inside the sphere and outside of the sphere.
    Part (b): Find the electric fields in these two cases. Show for the first case it is identical to a conducting sphere in an electric field.

    2. Relevant equations

    3. The attempt at a solution

    I have found potential inside and outside of sphere by solving the boundary conditions: ##\phi_in = \phi_out## and ##\epsilon_{r1} \epsilon_0 \frac{\partial \phi_{in}}{\partial r} - \epsilon_{r2} \epsilon_0 \frac{\partial \phi_{out}}{\partial r} = \sigma##:

    [tex]\phi_{in} = - \frac{\sigma_0 + 3\epsilon_0\epsilon_{r2}E_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} r cos \theta[/tex]

    [tex]\phi_{out} = \left[ -E_0 r + \frac{\left( E_0 \epsilon_0 (\epsilon_{r1} - \epsilon_{r2} \right) - \sigma_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} \frac{a^3}{r^2} \right] cos \theta [/tex]

    For ##\sigma_0 = 3\epsilon_0 E_0##:

    [tex]E_{in} = -\nabla \phi_{in} = \frac{\sigma_0 + 3\epsilon_0\epsilon_{r2}E_0}{\epsilon_0 (\epsilon_{r1} + 2\epsilon_{r2})} cos \theta[/tex]
    [tex]= \frac{3E_0 + 3\epsilon_{r2} E_0}{\epsilon_{r1} + 2\epsilon_{r2}} cos \theta [/tex]

    And for outside:
    [tex]E_{out} = \left[ 1 + \frac{2(\epsilon_{r1} - \epsilon_{r2} - 3)}{\epsilon_{r1} + 2\epsilon_{r2}} \left( \frac{a}{r}\right)^3 \right] E_0 cos \theta [/tex]

    I'm not sure what I'm supposed to look out for?

    For ##\sigma_0 = -(\epsilon_r -1)\epsilon_0 E_0##:

    I'm assuming outside is free space with ##\epsilon_{r2} = 1##:

    [tex]E_{in} = \frac{\sigma_0 + 3\epsilon_r E_0 \epsilon_0}{\epsilon_0 (\epsilon_r + 2} cos \theta [/tex]
    [tex] = \frac{1 - \epsilon_r + 3}{2 + \epsilon_r} E_0 cos\theta[/tex]
    [tex] = \frac{4 + \epsilon_r}{2 + \epsilon_r} E_0 cos \theta [/tex]

    For outside:
    [tex]E_{out} = \left[ E_0 + \frac{2\left( E_0 \epsilon_0 (\epsilon_r - 1) - \sigma_0 \right)}{\epsilon_0 (\epsilon_r + 2)} \left( \frac{a}{r}\right)^3 \right] cos \theta [/tex]
    [tex] = \left[ 1 + \frac{2(\epsilon_r - 1) + (\epsilon_r - 1)}{\epsilon_r + 2}\left( \frac{a}{r}\right)^3 \right] cos \theta [/tex]
    [tex] = \left[ 1 + \frac{3(\epsilon_r - 1)}{\epsilon_r + 2} \left( \frac{a}{r}\right)^3 \right] cos \theta [/tex]

    What's the deal with the conductor in an electric field?
    I know that for a conducting sphere, the positive charges migrate to +z while the negative charges migrate to -z and they are all spread on the surface of the sphere.
    Last edited: May 29, 2014
  2. jcsd
  3. May 30, 2014 #2
    I have searched and found online that when ##\sigma_0 = 3\epsilon_0 E_0##, the field inside would simply me ##-E_0 r cos \theta##.

    This clearly isn't the case here, i'm not sure why!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted