Calculation the resistance of a spherical shape

[green-fresh]

hello there!
I was reading about ohm rule in a MIT physics course and they calculate the resistance of a nested spherical shells like this :

and they but the microscopic form of ohm's law which is :
[tex]J=\sigma_q E[/tex]
and
[tex]I=A J[/tex]
so
[tex]I=(4 \pi r^2 ) (\sigma_q E)[/tex]
and
[tex]E=-\frac{\partial V}{\partial r}[/tex]
and he said that the potential must be like this form
[tex]V=\frac{C}{r}+D[/tex]
where : C and D are constant .
so my question is what is this form and why he did that and i could calculate the potential easily :
[tex]V_{ab}=\int E . dr[/tex].
thank you

 

[Petr Mugver]

For symmetry, the electric field is directed radially. Since there are no charges inside the conductor, the flux

[tex]\Phi=4\pi r^2 E[/tex]

must be constant. The difference of potential between a and b is

[tex]\Delta V=\int_a^b\frac{\Phi dr}{4\pi r^2}=\frac{\Phi(b-a)}{4\pi ab}[/tex].

But, as you said, [tex]I=\sigma\Phi[/tex], so

[tex]\Delta V=\frac{I(b-a)}{4\pi ab\sigma}\equiv RI[/tex]

and hence

[tex]R=\frac{(b-a)}{4\pi ab\sigma}[/tex]

 

[green-fresh]

thank you i think it is solved now .