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- Thread starter gambit1414
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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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