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Magnitude of electric flux through a sphere?

  1. Jan 30, 2017 #1
    ****EDIT****: I had improper significant figures. It was the correct number.
    1. The problem statement, all variables and given/known data

    The electric field on the surface of a 6.0 cm -diameter sphere is perpendicular to the surface of the sphere and has magnitude 52 kN/C .

    What is the magnitude of the electric flux through the sphere? Notice that the units of the answer involve kilo-newtons.

    2. Relevant equations
    I = EA or I = E4pir2 = Q/epsilon

    3. The attempt at a solution
    I first attempted to try and use Gauss's Law, but didn't get as far as I had hoped since we are not given a charge as thought previously.

    I then decided to us E*4pir2 to get 5881.06 kN *cm2/C which is 0.588106 kN*m2/C which is wrong. So I basically do not know how to approach this problem.

    Thanks in advance for any help!!!
  2. jcsd
  3. Jan 30, 2017 #2

    Charles Link

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    Homework Helper
    Gold Member

    The problem statement seems to be very incomplete because it doesn't say where the electrical charge is that is creating this electric field. You will get a different answer depending on the distribution of this charge. e.g. is it a surface charge, is it uniformly distributed throughout the sphere, or is there a point source at the center of the sphere?
    Last edited: Jan 30, 2017
  4. Jan 30, 2017 #3
    My answer was correct, but was missing a few significant figures. I think in general it was just wanting us to use Gauss's Law on the most simplified version possible. Indeed it is incomplete, but that was everything the question gave me unfortunately.
  5. Jan 30, 2017 #4


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    Staff: Mentor

    If the magnitude of the flux is uniform over the entire surface of the sphere then we can assume a spherically symmetric charge distribution that can be then be treated as a point charge located at the sphere's center.
    Strange, as I get quite different digits when I do the same calculation.
  6. Jan 30, 2017 #5
    Huh? Really? This is a screenshot of the question and answer.

  7. Jan 30, 2017 #6


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    Staff: Mentor

    My apologies. I mistook the diameter value to be a radius :oops: I hate when that happens :mad:

    Anyhow, it's great that you arrived at the correct result. Cheers!
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