Using the divergence theorem to prove Gauss's law?

AI Thread 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 a potential function f. The user initially believes that the divergence of E is zero, leading to confusion in their calculations. After some back and forth, they realize they made an error in calculating the divergence of E. The conversation highlights the importance of correctly applying vector calculus in the context of electromagnetic theory. Ultimately, the user acknowledges their mistake and expresses gratitude for the clarification.
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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