Griffith's EM third ed. pp.68 equ 2.12 ....(adsbygoogle = window.adsbygoogle || []).push({});

"In the case of a point charge q at the origiin, the flux of E though a sphere of radius r is

[tex]\int \vec{E} \cdot d\vec{A} = \int \frac{q}{4\pi\epsilon_{0}r^2}\hat{r} \cdot (r^2sin\theta \ d\theta \ d\phi \ \hat{r})[/tex]"

How the hell is dA equal to that? Multiply dr to it and it's dV in spherical coord. The area of a sphere is 4 pi r^2. Shouldn't dA be 8 pi r dr?

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# Griffith's EM third ed. pp.68 equ 2.12

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