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Gauss's law, nucleus, potential proton due to other protons

  • Thread starter binbagsss
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  • #1
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I'm a little out of touch with this stuff , but I'm really not getting it..

So my book is considering: in a nucleus of ##z## protons, consider one proton in spherical charge distribution to other protons.

So ## \rho = (z-1)e/(4/3) \pi R^{3}##, where ##R## is the radius of the nucleus, is the charge density. (this is fine)

I have for Gauss's law:

##\int E.d\vec{A} = Q_{enclosed}/\Sigma_{0}##,

So I get ##E.4\pi r^{2}=\rho (4/3) \pi r^{3} / \Sigma_{0}##
##=>## ##E=(z-1)e r/4\pi R^{3}##

Whereas my book has ##q/4 \pi \Sigma_{0} r= (4/3) \pi r^{3} \rho/4 \pi \Sigma_{0}r=(z-1)e/4 \pi \Sigma_{0}(r^{2}/R^{3}##,

I dont understand the far LHS of this? But anyway I'm a factor ##r## out, I have no idea how the book gets ##r^{2}##,

your assistance is greatly appreciated, thank you !
 
Last edited:

Answers and Replies

  • #2
BvU
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( I hope it's OK with you if I use ##\varepsilon_o## for the permittivity of the vacuum. With your symbol, I keep wondering what gets summed :wink:)

$${1\over 4 \pi \varepsilon_o}\; {q\over r} $$ looks more like a potential, not an electric field strength. That would explain the factor r.
 

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