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Clarification on Electric Flux

  1. Feb 18, 2009 #1
    A question has 2 spheres of different diameters surrounding equal charges 'q'. Diameter of sphere A is smaller than diameter of sphere B. Now are the flux for both equal or not? I think they are equal because total flux = Qenclose/ epsilon-knot, so flux only depends on the charges. But then again the electric field on the bigger sphere (B) would be weaker than the E-field on sphere A examining the equation for E-field and flux is equal to the surface integral of E dot ds. So i'm questioning which is the correct answer. Thank You.
     
  2. jcsd
  3. Feb 18, 2009 #2
    Let us take a look at some equations. That should clear up your question:

    The surface integral of the flux (D) out of a closed surface (S) is equal to the charge (Q) contained within the surface (Gauss' Law):
    [tex]

    \int \int_s \vec{D} d \vec{S} = Q

    [/tex]

    This can be solved for D as follows if the surface is a sphere where r is the radius of the sphere:

    [tex]

    \vec{D} = \frac{Q}{4\pi r^2} \vec{a}

    [/tex]

    Now look at equation 2. What affect does r have on the magnitude of the flux?
     
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