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Gauss's Law with a Insulating Shell

  1. Sep 13, 2014 #1
    1. The problem statement, all variables and given/known data
    pic.gif
    An insulator in the shape of a spherical shell is shown in cross-section above. The insulator is defined by an inner radius a = 4 cm and an outer radius b = 6 cm and carries a total charge of Q = + 9 μC(You may assume that the charge is distributed uniformly throughout the volume of the insulator).

    What is Ey, the y-component of the electric field at point P which is located at (x,y) = (0, -5 cm) as shown in the diagram?

    2. Relevant equations
    $$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$


    3. The attempt at a solution
    I'm confused on where the boundries of the shell lie, is it between the end of a and the end of b (thus making the shell have a width of b-a)?

    If it IS the above, then we'd have to do analyze the problem with a Gaussian surface that ends at point P, located between the two.

    We already know:
    $$\rho = \frac{+Q}{\frac{4}{3}\pi r^3}$$​

    $$E\oint dA = \frac{Q_{enclosed}}{\epsilon_0}$$​
    $$EA = \frac{Q_{enclosed}}{\epsilon_0}$$​
    $$E(4\pi r^2) = \frac{Q_{enclosed}}{\epsilon_0}$$​
    $$E(4\pi r^2) = \frac{\rho \frac{4}{3}\pi r^3}{\epsilon_0}$$​
    $$E = \frac{\rho r}{3\epsilon_0}$$​

    However, when I attempt to use the above expression, I'm not getting a correct answer. What am I doing incorrectly, I'm definitely confused on what "r" should I be using (i.e. is "r" the .05 m given to us in the problem or is, for some reason, the sphere with radius P minus the sphere with a radius of a)?
     
  2. jcsd
  3. Sep 13, 2014 #2

    mfb

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    2016 Award

    Staff: Mentor

    Right.

    That is an equation for a uniform sphere, not for a spherical shell.
    What is the volume of the spherical shell?

    You'll need another volume of a (different) spherical shell for the electric field afterwards.
     
  4. Sep 14, 2014 #3
    Thank you!

    I understood my error basically as soon as I wrote out the question despite struggling for a while.

    For anyone else with a similar problem looking at this, basically there's only charge in the "shell" between b and a. So to find the charge density you'd have to do the equation for a shell with radius b and subtract from that the shell with radius a, basically netting you with:

    $$\rho = \frac{Q}{\frac{4}{3}\pi (b^3-a^3)}$$​
     
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