Grade 12Organic Chemistry

the flux of a constant vector through any closed surface is zero.. can you explain the statement with an example?

Gautam Mishra

6 Years agoGrade 12

3 Answers

Arun

6 Years ago

am assuming , you meant flux through any closed surface of any shape or any area, over Vector Field is zero. Looks to me like a Vector Field with Zero Divergence at every point. Limit definition of Divergence at a point is

$Div F =lt_{V↦0} \frac{\oint _A \vec{F} .\vec{dA}}{V} $

$=lt_{V↦0} \frac{\Phi _A}{V}$

In the above formula , it's meant to be double integral , but couldn't find the latex syntax . The syntax $\oiint$ didn't work.

Note here, $\Phi_A $ is the total flux through the closed surface , and not any surface. Which by your question is Zero.

So it's a field with zero divergence at any point. Zero divergence through any closed surface implies Vector Field with zero divergence everywhere.

Now it doesn't mean that , flux through some random surface that is not closed , be zero as well. Flux through the surface of sphere is by assumption zero, but there may still be flux through some sheet of paper in space. And here comes the world of Magnetic Field. A solenoidal magnetic field with zero divergence everywhere - no source or sink.

Examples of Solenoidal Field in 3D is difficult to find,so I will give you simplest of example.

$F = (xy)\hat{j}-(xz)\hat{k}$

$DivF=∇.F = 0$

⇒ Flux through any closed Surface is also Zero. Use divergence theorem

$\oint _A \vec{F}.\vec{dA} =∫∫∫_V divF dV$

Vikas TU

6 Years ago

Dear student

I don’t think that’s true. Consider a uniform vector field. It is both solenoidal (zero divergence, flux through any closed surface) and irrotational (zero curl).

If the vector field (in 3D space) decays more rapidly than 1/r to zero as you approach infinity and the field is “sufficiently smooth” (true of all physical fields), then Helmholtz’s decomposition theorem applies.
nd the only way for both the divergence and curl to be zero is for the vector field to be identically zero. In that case, you can’t have a non-zero vector field for which the flux through all closed surfaces (divergence) is zero and the curl is everywhere zero.

Physically, there are no uniform vector fields throughout all space if the field is created by sources. That is probably what the statement you are thinking about is getting at.

Hope this helps and does not mislead or confuse you.