Why is the flux of a conservative field always positive?

Question

I have been given this question as a task. I have done some mathematical analysis but they were of no help. The only thing I could think of was that total flux in a closed surface can be calculated by the divergence theorem but since we don't really have a specific equation for the surface or even the field how would I prove this then?

Answer 1

The flux of a vector field through a surface is a signed quantity, and is multiplied by $-1$ when you change the orientation (normal direction) of the surface.

If $S$ is a closed surface then one might assume it a priori outward oriented, by general agreement. But even then the statement in the title of your question is wrong. Consider the function $f(x,y):=1-x^2-y^2$ in the plane. Its gradient $\nabla f(x,y)=(-2x,-2y)$ is a conservative vector field. The flux of this field through the outward oriented boundary $\partial D$ of the unit disc $D$ is obviously negative.

(Of course there is the analogous example in three dimensions. Just define $f(x,y,z):=1-x^2-y^2-z^2$, and replace the disc $D$ by the unit ball $B\subset{\mathbb R}^3$.)