Using the divergence theorem to prove Gauss's law?

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SUMMARY

The discussion centers on using the divergence theorem to prove Gauss's law by computing the integral of the electric field E derived from the potential function f(X) = q/(4πr), where r = ||X||. The user initially misapplied the divergence operator, leading to the incorrect conclusion that div E = 0. Upon reevaluation, the user identified an error in their calculations, confirming that the divergence of E is not zero, thus aligning with Gauss's law.

PREREQUISITES
  • Understanding of vector calculus, specifically divergence and gradient operations.
  • Familiarity with Gauss's law in electrostatics.
  • Knowledge of parametrization of surfaces, particularly spheres.
  • Proficiency in using the divergence theorem in three-dimensional space.
NEXT STEPS
  • Study the application of the divergence theorem in various coordinate systems.
  • Explore detailed examples of Gauss's law in electrostatics.
  • Learn how to compute divergences of vector fields in Cartesian coordinates.
  • Investigate common mistakes in vector calculus and how to avoid them.
USEFUL FOR

This discussion is beneficial for physics students, mathematicians, and educators focusing on electromagnetism and vector calculus, particularly those looking to deepen their understanding of Gauss's law and the divergence theorem.

kittyset
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Hello,

I've been struggling with this question:

Let q be a constant, and let f(X) = f(x,y,z) = q/(4pi*r) where r = ||X||. Compute the integral of E = - grad f over a sphere centered at the origin to find q.

I parametrized the sphere using phi and theta, crossed the partials, and got q, but I think there's another way using the divergence theorem, given as ∫∫E⋅ndσ = ∫∫∫ div E dV (sorry about the awkward symbol usage :/ ). I'm not sure what's going wrong with the following:

1. div grad f = div -E = div E = 0

2. ∫∫∫ div E dV = 0 ≠ q

I'm probably missing something super basic, but any hint would be a great help!
 
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Why do you say that div E = 0?
E = -grad f = ##-\begin{pmatrix} \frac{\partial }{\partial x } &\frac{\partial }{\partial y }& \frac{\partial }{\partial z }\end{pmatrix} \frac{q}{4\pi \sqrt{ x^2 + y^2 +z^2 } }. ##
This is not a constant, so why would the divergence be zero?
 
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RUber said:
Why do you say that div E = 0?
E = -grad f = ##-\begin{pmatrix} \frac{\partial }{\partial x } &\frac{\partial }{\partial y }& \frac{\partial }{\partial z }\end{pmatrix} \frac{q}{4\pi \sqrt{ x^2 + y^2 +z^2 } }. ##
This is not a constant, so why would the divergence be zero?

Whoops, you're totally right. Found an error when i took the div. Thanks so much!
 

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