- #1

- 1,232

- 10

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}##,

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: