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Homework Help: Spherical Shells and Gauss' Law

  1. Feb 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A small, insulating, spherical shell with inner radius a and outer radius b is concentric with a larger insulating spherical shell with inner radius c and outer radius d. The inner shell has total charge q distributed uniformly over its volume, and the outer shell has charge -q distributed uniformly over its volume.


    Calculate the magnitude of the electric field and direction of the field (outward or toward the center) for

    i. a < r < b
    ii. b < r < c
    iii. c < r < d

    2. Relevant equations

    See below.

    3. The attempt at a solution
    General workings for shell with inner radius a and outer radius b:

    Q = rho*(4/3)*pi[b^3 – a^3], where Q = total uniform charge

    indefinite integral[E*dA] = q_enclosed/epsilon_0

    E*4*pi*r^2 = [rho*(4/3)*pi[r^3 – a^3]]/[epsilon_0]

    Note that then rho for inner shell = q/[(4/3)*pi*(b^3-a^3)] and outer shell rho = [-q]/[(4/3)*pi*(d^3-c^3)] in this specific problem.

    a. E = [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2)

    The direction will be away from the center??

    b. Total charge is q, so E = (1/4*pi*epsilon_0)*(q/r^2) This will be outward the center, too?

    c. This one I am really unsure of—both with the direction and electric field expression since two different charges are featured.
    At first, I thought it would be [-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)], but this alone is wrong???

    Would it be a sum:

    [-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)] + [q/(4*pi*epsilon_0)]*[(r^3 – a^3)/(b^3-a^3)]*(1/r^2) + (1/4*pi*epsilon_0)*(q/r^2) ????

    As for direction, would it be outward, too? A positive test charge would move away from the positively charged shell and move towards the negative shell?

    Any help is appreciated. Thank you.
    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 3, 2007 #2
    Can anyone please help me with part iii. and check my direction answers for the other parts?

    Thank you again.
  4. Feb 3, 2007 #3

    Doc Al

    User Avatar

    Staff: Mentor

    These look OK to me.

    Here it seems like you just considered the outer shell of charge but forgot to include the inner shell.

    It would be the sum of the field from the partial outer shell plus the field from the total inner shell--you've computed them both, just add them up.

    The field at point r just depends on the total charge within the spherical surface at r. In this problem, that total charge is always positive for all points up to r = d, thus it points outward at all points.
  5. Feb 3, 2007 #4
    The sum will be:

    total inner shell, (1/4*pi*epsilon_0)*(q/r^2), + partial outer shell,
    [-q*(r^3 – d^3)]/[4*pi*epsilon*r^2*(c^3 – d^3)] ?

    Or did I use/mismatch the wrong expressions?
  6. Feb 3, 2007 #5

    Doc Al

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    Staff: Mentor

    That looks good to me.
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