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Gaussian Surface and Charge Density

  1. Jul 30, 2013 #1
    1. A solid dielectric sphere of radius 10 cm has an electric charge uniformly distributed throughout its volume. The electric field at 5 cm from the center of the sphere is 8.6 x 10^4 N/C, pointing radially outward. Calculate the magnitude and direction of the electric field at a point 15 cm from the center of the sphere.

    2. Gaussian Law


    3. I solved it like this:

    E(4∏r2) = Q(r3/a3)/ ε0
    E = Qr/4∏ε0a3
    Q = 1.91 * 10^-7C
    E = kQ/r2
    = 76.4kN/C radically outward.

    My question is what is the difference between doing it the above way versus with charge density:
    Finding r < a (inside the sphere)
    Equation: E(4∏r2) = 4/3(∏r3)ρ/ ε0
    ρ = 4.5666 *10^-5

    Find r > a (outside the sphere)
    E(4∏r2) = 4/3(∏a3)ρ/ ε0
    E = 76.4 kN/C radically outward
     
  2. jcsd
  3. Jul 30, 2013 #2

    Dick

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    None really. If you use ρ=Q/((4/3)*pi*a^3) (charge/volume) and substitute into your second equation for E(4*pi*r^2) you get the same formula as you used in the first part. The only difference is whether you choose to solve for Q first or ρ first.
     
    Last edited: Jul 30, 2013
  4. Jul 31, 2013 #3
    Thank You!!
     
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