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I have the vector field w.r.t the spherical coordinates $(r,\theta,\phi)$ : $\mathbf{F}=r^2\cos(\phi) \mathbf{e}_{\theta}$

why the flux through the sphere is zero ?

Thanks you very much.

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Your field looks scalar, yet it should be vector. –  Ron Gordon Mar 19 '13 at 12:02
    
Do you know the definition of the flux through a surface? And have a look at your vector field again; it doesn't seem to be well-defined everywhere. –  Rhys Mar 19 '13 at 12:10
    
exactly it does not defined any where , but this is the question ,we have just this formula and we were asked to give an explanation why the flux is 0 –  user2157473 Mar 19 '13 at 12:12

2 Answers 2

up vote 2 down vote accepted

Your field is

$$\vec{F} = r^2 \cos{\phi} \vec{e_{\theta}}$$

As there is only a $\theta$ component of the field, then the divergence of this field is

$$\vec{\nabla}\cdot \vec{F} = \frac{1}{r \sin{\theta}} \frac{\partial}{\partial \theta} (\sin{\theta} F_{\theta}) = r \cos{\phi} \cot{\theta}$$

The net flux through the sphere is the integral of the divergence of the field through the volume of the sphere:

$$\int_0^R dr \: r^3 \int_0^{\pi} d\theta \: \cos{\theta} \int_0^{2 \pi} d\phi \: \cos{\phi}$$

This integral, however, is zero because the integral over $\phi$ is zero.

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That's slight overkill. $\vec{e_\theta}\cdot\vec{e_r} = 0$. –  Rhys Mar 19 '13 at 12:17
    
Thanks a lot for this nice explanation –  user2157473 Mar 19 '13 at 12:18
1  
@Rhys: for you, maybe. Not for the OP. –  Ron Gordon Mar 19 '13 at 12:18
    
Well, the definition of the flux through a surface is not a volume integral. What if they were asked to calculate the flux through the northern hemisphere? –  Rhys Mar 19 '13 at 12:20
    
actually the both answers are useful for me , because at first they asked us to give the reason why it is 0 , and then by divergence –  user2157473 Mar 19 '13 at 12:21

It is known as Gauss's Law and it holds for any closed surface.

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Can you be a little more specific? –  Ron Gordon Mar 19 '13 at 12:02
    
No it doesn't; that's only for divergence-free vector fields. –  Rhys Mar 19 '13 at 12:07
    
it is a vector but I do not know how to put e(theta) like a symbol So I mean just like 3i+4j+5k and here it is just on the e(theat) –  user2157473 Mar 19 '13 at 12:07
    
So "e(theta)" is $e_{theta}$? Knowing that then, I see that this field is not divergence-free, but the integral about its aziumuthal angle of the divergence is zero. –  Ron Gordon Mar 19 '13 at 12:10
    
Thank you very much for helping me –  user2157473 Mar 19 '13 at 12:16

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