Flux of vector field proportional to 1/r^3 through sphere

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SUMMARY

The discussion focuses on calculating the flux of the vector field \(\vec{F} = \frac{\vec{r}}{r^3}\) through a sphere of radius "a" centered at the origin. The Gauss Divergence Theorem is referenced, but the divergence \(\nabla \cdot \vec{F}\) is found to be zero, leading to confusion regarding the flux calculation. The key conclusion is that due to the singularity at the origin, the divergence theorem cannot be applied directly, necessitating a direct computation of the flux instead.

PREREQUISITES
  • Understanding of vector fields and their properties
  • Familiarity with the Gauss Divergence Theorem
  • Knowledge of spherical coordinates for flux calculations
  • Concept of singularities in vector fields
NEXT STEPS
  • Learn how to compute flux directly from the definition of a vector field
  • Study the implications of singularities in vector calculus
  • Explore alternative forms of the Gauss Divergence Theorem for singular fields
  • Review spherical coordinate transformations for vector field analysis
USEFUL FOR

Students and professionals in physics and engineering, particularly those studying electromagnetism and fluid dynamics, will benefit from this discussion on vector field flux calculations.

jameson2
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Homework Statement


Consider the vector field [tex]\vec F=\frac{\vec r}{r^3}[/tex] with [tex]\vec r=x\hat{i}+y\hat{j}+z\hat{k}[/tex]. Compute the flux of [tex]\vec F[/tex] out of a sphere of radius "a" centred
at the origin.

Homework Equations


The Gauss Divergence Theorem [tex]\int_D dV \nabla \bullet F=\int_S F\bullet dA[/tex]


The Attempt at a Solution


I think I'm either missing or not understanding something in this question. When I compute [tex]\nabla \bullet F[/tex], I get zero, which means the flux is zero, I think. But this doesn't seem right at all. What am I missing?
 
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The vector field [tex]\vec{F}[/tex] has a singularity at the origin, so you can't use the divergence theorem (at least not in its most common form). Compute the flux directly from the definition instead.
 

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