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Homework Help: Charged Sphere with Cavity

  1. Apr 22, 2008 #1
    An insulating sphere of radius 5.00 cm , centered at the origin, has a uniform volume charge density 3.05 uC/m^3 . There is a spherical cavity cut out of its center of radius 2.00 cm.

    Well i know to use Gauss' Law:
    Φ = ∫E∙dA = Q/ε
    where Φ is the electric flux, E is the electric field, dA is a differential area on the closed surface S with an outward facing surface normal defining its direction, Q is the charge enclosed by the surface, and ε is the electric constant.

    1.) What is the electric field at 1.43 cm ?
    I found that to be zero.

    2.) What is the electric field inside the spherical shell at 2.86 cm ?

    Q = (4π/3)*(0.0286^3-0.0200^3)*3.05x10^-6C
    The surface area integrated over is: S = 4π*0.0286^2. Hence:
    E = Q/εS = (0.0286^3-0.0200^3)*3.05x10^-6/(3ε*0.028...

    for surface area i get 1.03*10^-2, for Q i get 4.188*1.53*10^-5*3.05*10^-6= 1.95*10^-10
    Then i divide Q/S which is 1.90*10^-12. but i keep getting it wrong. what am i doing wrong?

    3.) What is the electric field outside the spherical shell at 6.75 cm ?
    Q = (4π/3)*(0.0500^3-0.0200^3)*3.05x10^-6C
    The surface area integrated over is: S = 4π*0.0675^2. Hence:
    E = Q/εS = (0.0500^3-0.0200^3)*3.05x10^-6/(3ε*0.067...

    I did the same as part two equations and i get 3.89*10^-9 which is wrong also. Help!

    3. The attempt at a solution[/
  2. jcsd
  3. Apr 22, 2008 #2
    1) is right, since there is no charge in the cavity.

    For the others, I'm getting a headache reading all the digits :p It might be easier to spot a mistake if you write it out in symbols. I'll try that now.
  4. Apr 22, 2008 #3

    2) After using gauss' law you should find:
    [tex]E = \frac{Q}{\epsilon A}[/tex]
    Q is in this case dependend on the charge density rho and the volume V:
    [tex]Q = \rho V = \rho \frac{4 \pi}{3}(r^3 - {r_0}^3)[/tex] where r is 2.86 cm and r_0 is the radius of the cavity (2 cm)

    [tex]E = \frac{\rho \frac{4 \pi}{3}(r^3 - {r_0}^3)}{\epsilon 4 \pi r^2} = \frac{\rho (r^3 - {r_0}^3)}{3 \epsilon r^2}[/tex]

    Entering values:
    [tex]r = 0.0286, r_0 = 0.02, \epsilon = 8.85 \times 10^{-12}, \rho = 3.05 \times 10^{-6}[/tex] yields:
    [tex]E = 2161.9[/tex] (provided I didn't make any errors while entering the values in my calculator :p )

    What should the answer be according to you?
  5. Apr 22, 2008 #4
    Thanks A Bunch!
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