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Clarification on Electric Flux

  • Thread starter gambit1414
  • Start date
  • #1
12
0
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.
 

Answers and Replies

  • #2
27
3
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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