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Homework Help: Calculate E inside a spherical shell

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the electric field, E, for general points inside a shell which has a radius R. It also has a uniform surface charge density, σ.

    First, do it with Gauss’s Law.

    Then find E more directly. Hint: Use Law of Cosines and a substitution.

    2. Relevant equations

    3. The attempt at a solution

    I believe for the first part, the answer is just 0 since Q enclosed is 0.

    The second part is where I need more help.
  2. jcsd
  3. Sep 22, 2010 #2
    you are right for the first part.

    For the second part, what is your working thus far?
    How do you express the electric field at a point within the shell due to a point charge on the shell?
  4. Sep 22, 2010 #3
    I am trying to use spherical coords. Therefore, an arbitrary point inside of the shell sould be (r,theta,phi) where r<R. I am trying to get script r (for the denominator of the equation for E).
    Would that be (r-R, 0, 0). This is my trouble spot.
  5. Sep 22, 2010 #4
    here's a hint for you:

    Place the point in the shell for which you are trying to calculate the electric field on the z-axis. You should be able to see how to use the cosine rule from there. You can also easily switch back to using spherical coordinates for doing the integration.
  6. Sep 22, 2010 #5
    The my script r becomes....
    (script r)^2= R^2 +z^2 -2Rzcos(theta')

    Will this still help me for points not on the z axis. If so, I still am not sure where to go.
  7. Sep 22, 2010 #6
    you see, you are calculating for any arbitrary point in the shell. The point remains stationary when you do the integration. So you are free align your z-axis however you want it. Arbitrariness is not lost since you are free to align your z-axis.

    By the way, your equation is correct. Now all you have to do is the integration.
  8. Sep 22, 2010 #7
    For the integration, I can pull sigma out because it's constant. That leaves me with...

    r-hat R^2 sin(theta')d(theta')d(phi') / R^2 +z^2 -2Rzcos(theta')

    My bounds for d(theta') are 0 to Pi and my bounds for d(phi') are 0 to 2Pi?

    Do I need to convert r-hat?
  9. Sep 22, 2010 #8
    yes. Notice that for every loop around the sphere when you do the integration for phi, the horizontal components cancel. Now all you have to do is to add up the vertical components i.e. the components that run along the z-axis.
  10. Sep 22, 2010 #9
    So would it be rcos(theta) that is added onto my integral instead of r-hat?
  11. Sep 22, 2010 #10
    no. Look closely at your diagram. you need the horizontal component of the electric field vector.
  12. Sep 22, 2010 #11
    I thought the horizontal components cancelled and I needed the vertical ones?
  13. Sep 22, 2010 #12
    oh yeh right the vertical ones i meant. you can find cos alpha, where alpha is the angle between the vertical and the electric field vector by considering the lengths of the triangle.
  14. Sep 22, 2010 #13
    would sin alpha be R/z ?
  15. Sep 22, 2010 #14
    No, cos alpha should be

    [tex] \frac{R\cos{\theta}-z}{(R^2+z^2-2Rz\cos{\theta})^{(\frac{1}{2})}}[/tex]

    the denominator is to the power of half, and the numerator is R costheta instead of R, the latex needs a while before it refreshes.
    Last edited: Sep 22, 2010
  16. Sep 22, 2010 #15
    I think I am visualizing alpha and it's corresponding segments wrong.
  17. Sep 22, 2010 #16
    That's just the definition for law of cosines. I don't know how to proceed from here though.
  18. Sep 23, 2010 #17
    do you have a diagram to show? have you drawn your diagram correctly?
  19. Sep 23, 2010 #18
    unfortunately my scanner sin't working properly right now and the assignment is due in an hour
  20. Sep 23, 2010 #19
    well here is a hand drawn diagram using paint.

    [PLAIN]http://img691.imageshack.us/img691/3497/32804960.jpg [Broken]
    Last edited by a moderator: May 4, 2017
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