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Electric field inside a cavity

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    If I have a solid sphere of radius R and charge density [itex]+\rho \, C/m^3[/itex] and I then remove a smaller sphere of radius [itex]b[/itex] and is a distance [itex]a[/itex] from the center of the larger sphere, what is the electric field inside the cavity?
    I get an answer which I think is right. I've look at the math over and over and can't quite figure out what the teacher did.

    2. Relevant equations
    Gauss' Law:
    [tex] \oint \vec{E} \cdot d\vec{S} = \frac{q_{encl}}{\epsilon_0} [/tex]



    3. The attempt at a solution
    I know this is a classic problem using the superposition principle.
    First I apply GL to the large sphere at a distance [itex]r<R[/itex] from the center and get an electric field
    [tex] \vec{E} = \frac{ \rho }{3 \epsilon_0} \vec{r} [/tex]

    Then I do the same except with a charge density of [itex]-\rho \, C/m^3[/itex] so
    [tex] \vec{E'} = - \frac{\rho }{3 \epsilon_0} \vec{r'} = -\frac{\rho}{3 \epsilon_0} (\vec{r} - \vec{a}) [/tex]

    Now I just sum the two fields
    [tex] \begin{align}
    \vec{E} + \vec{E'} & = \frac{ \rho }{3 \epsilon_0} \vec{r} -\frac{\rho}{3 \epsilon_0} (\vec{r} - \vec{a}) \\ &= \frac{ \rho }{3 \epsilon_0} \vec{r} -\frac{\rho}{3 \epsilon_0} \vec{r} +\frac{\rho}{3 \epsilon_0} \vec{a} \\ &=\frac{\rho}{3 \epsilon_0} \vec{a} \end{align} [/tex]

    This is my solution which is supposedly wrong.
     
  2. jcsd
  3. Oct 28, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your solution is correct. See, for example http://jkwiens.com/2007/10/24/answe...nonconducting-sphere-with-a-spherical-cavity/
    What did your teacher do?

    ehild
     
  4. Oct 28, 2012 #3
    wrong thread :P
     
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