# Calculation the resistance of a spherical shape

1. Aug 18, 2010

### 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 :
$$J=\sigma_q E$$
and
$$I=A J$$
so
$$I=(4 \pi r^2 ) (\sigma_q E)$$
and
$$E=-\frac{\partial V}{\partial r}$$
and he said that the potential must be like this form
$$V=\frac{C}{r}+D$$
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 :
$$V_{ab}=\int E . dr$$.
thank you

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2. Aug 18, 2010

### Petr Mugver

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

$$\Phi=4\pi r^2 E$$

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

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

But, as you said, $$I=\sigma\Phi$$, so

$$\Delta V=\frac{I(b-a)}{4\pi ab\sigma}\equiv RI$$

and hence

$$R=\frac{(b-a)}{4\pi ab\sigma}$$

3. Aug 18, 2010

### green-fresh

thank you i think it is solved now .