# What is the resistance of an ohmic sphere with surface potential $\phi_0\cos\theta$?

I'm trying to find the volumetric density of current in the interior of a sphere (made of ohmic material) with radius $a$, conductivity $g$ and surface potential $\phi=\phi_0cos\theta$.

This is the way I thought I could solve it:

First, it seems to me that the electric field should have only a $z$ component since the potential is symmetric with respect to the azimutal angle and increases when we move along the $z$ axis, having values $\phi=-\phi_0$ when $z=-a$ and $\phi=\phi_0$ when $z=a$. (Is this assumption correct?)
Secondly, I would compute the value of the resistance $R$ of the sphere using the formula $R=\frac{\Delta L}{gS}$ for a cylindric conductor of section $S$ and length $\Delta L$. In order to do that I have to sum up the resistance of all the cylinders of infinitesimal lenght $dz$, section $S(z)=\pi(a^2-z^2)$ and resistance $dR(z)=\frac{dz}{\pi g(a^2-z^2)}$. So, I would have: $$R=\int_{-a}^a\frac{dz}{\pi g(a^2-z^2)}$$ Then I would use Omh's law fo find $I$ (using that $\Delta\phi=2\phi_0$), and finally I would obtain the volumetric density of current $J$ as a function of $z$ using the formula $J=\frac{I}{S(z)}$.

Now, the problem is that the integral for the resiastance diverges, which would imply $J=0$ and it doesn't seem to be correct. What is wrong with my approach?

• Why the downvote? It would be much more helpful to give me a suggestion on how to improve my question. Commented Oct 9, 2016 at 23:21
• Comments are not for extended discussion; this conversation has been moved to chat. Commented Oct 10, 2016 at 4:22

You can't really solve this problem using $z$-slices, disks, because not all of the current passes through every disk. If all of the current had to pass through every disk then the resistance will be infinite because of the need to pass finite current through the points with zero cross sectional area at opposite sides of the sphere. Instead, you'll need to add cylindrical shell conductors in parallel. When resistors are in parallel, the conductance, $1/R$, adds. The conductance of a cylindrical shell with height $2z$, radius $r$, and width $\operatorname{d}r$ is: $$\operatorname{d}\left(\frac{1}{R}\right) = \frac{g 2\pi r \operatorname{d}r}{2z}.$$ Because the cylindars are contained in a sphere, the half-height, $z$ is given by the equation: $$z = \sqrt{a^2 - r^2}.$$ So the total conductance is: \begin{align}\frac{1}{R} &= \pi g\int_0^a \frac{r}{\sqrt{a^2 - r^2}}\operatorname{d}r \Rightarrow \\ R &= \frac{1}{\pi g a}. \end{align} This, at least, will be the resistance if $\Phi(r=a) = \phi_0 \operatorname{sgn}(z)$ on the surface of the sphere. For your situation the need to use $\mathbf{J} = \sigma \mathbf{E}$ may be unavoidable.
• Right, and I added the caveat in the last sentence. The total resistance is likely taken as $2\phi_0 / I_{\mathrm{tot}}$, with $I_{\mathrm{tot}}$ calculated from $\mathbf{J}$ from the continuous version of Ohm's law. Commented Oct 10, 2016 at 0:39
• If $E=\phi_0/a$, then $J = g\phi_0/a$. Thus $I_{\mathrm{tot}} = \pi g \phi_0 a$. If we define $R = 2\phi_0 / I_{\mathrm{tot}} = 2 / (\pi g a).$ The reason for using $V=2\phi_0$ in Ohm's law is because that's the difference between the top and bottom of the sphere. Commented Oct 10, 2016 at 0:49