# Electrical Resistance of a Sphere?

1. Nov 21, 2012

### greswd

How does one find the electrical resistance of a homogenous sphere of uniform density?

By connecting two wires across the diameter of the sphere, and assuming Pouillet's Law $$R=ρ\frac{L}{A}$$ holds.

Last edited: Nov 21, 2012
2. Nov 21, 2012

### Staff: Mentor

I don't think this is well-defined, if you try to calculate the resistance between two points on the sphere. If you have some finite area where you connect electrodes, you get some finite resistance value, but I think you will need a numerical simulation to calculate it.

3. Nov 21, 2012

### cabraham

You need to truncate the sphere by lopping off a small section at each end. That way the electrodes have some area. If a flat planar electrode pair is pressed against opposite sides of a true sphere, the resistance is infinite since the surface area between the sphere contact & electrode is literally zero. If the sphere ends are flattened, a finite area exists, & the resistance is finite.

Integration will compute the actual resistance. If a sphere is 10 cm in diameter, & the opposite ends of the sphere are lopped off resulting in 1.0 cm diameter contact surface area, you can compute resistance from those values. Did I help?

Claude

4. Nov 21, 2012

### marcusl

Claude, that is an approximation. Ut not exact because the material you excluded is in parallel and will result in lowering your R value. Conformal mapping might work for planar case (R of disk) followed by pi rotation. Numerical simulation will definitely work.

5. Nov 21, 2012

### greswd

But aren't the truncated parts in series, not parallel?

I tried numerical approximation but it leads to contradictory results. (See below)

Last edited: Nov 21, 2012
6. Nov 21, 2012

### greswd

First up, I have a solid sphere of radius $$r_0$$

I approximated by creating cylinders that "fill up" the sphere from inside, hence it is an under-approximation.

Let's look at one hemisphere. I split up the hemisphere into q cylinders of the same height.
$$q\inℤ^{+}$$

The height of each cylinder:
$$h=\frac{r_0}{q}$$

The radius of each cylinder, rn
$$(r_n)^{2}+(r_x)^{2}=(r_0)^{2}$$
$$(r_n)^{2}+(nh)^{2}=(r_0)^{2}$$

$$(r_n)^{2}=(r_0)^{2}-(nh)^{2}$$
$$(r_n)^{2}=(r_0)^{2}-n^{2}\frac{(r_0)^{2}}{q^{2}}$$
$$(r_n)^{2}=(r_0)^{2}\left[\frac{q^{2}-n^{2}}{q^{2}}\right]$$

The area of each cylinder
$$A_n=π(r_0)^{2}\left[\frac{q^{2}-n^{2}}{q^{2}}\right]$$

Total Resistance (ignoring the last cylinder):

$$R_T=ρ\frac{h}{A_1}+ρ\frac{h}{A_2}+ρ\frac{h}{A_3}+....+ρ\frac{h}{A_{q-1}}$$
$$R_T=ρh×\sum_{n=1}^{q-1} \frac{1}{A_n}$$
$$\frac{1}{A_n}=\frac{1}{π(r_0)^{2}}\left[\frac{q^{2}}{q^{2}-n^{2}}\right]$$
$$R_T=\frac{ρh}{π(r_0)^{2}}×\sum_{n=1}^{q-1} \left[\frac{q^{2}}{q^{2}-n^{2}}\right] =\frac{ρ\frac{r_0}{q}}{π(r_0)^{2}}×\sum_{n=1}^{q-1} \left[\frac{q^{2}}{q^{2}-n^{2}}\right] =\frac{ρ}{πr_0}×\frac{1}{q}\sum_{n=1}^{q-1} \left[\frac{q^{2}}{q^{2}-n^{2}}\right]$$

$$R_T=\frac{ρ}{πr_0}×\frac{1}{q}\sum_{n=1}^{q-1} \left[\frac{q^{2}}{q^{2}-n^{2}}\right] , q→∞$$

That should be the resistance of a hemisphere. The problem is,

$$\frac{1}{q}\sum_{n=1}^{q-1} \left[\frac{q^{2}}{q^{2}-n^{2}}\right] , q→∞$$

does not converge to a single value. It apparently keeps on increasing as q inccreases. Hence, it would seem that a sphere can have any amount of resistance. This is quite contradictory.

Last edited: Nov 21, 2012
7. Nov 21, 2012

### marcusl

This implies that the potential present at the electrode is also present at the top of each cylinder, and likewise for the bottom--but that would only be true if the surface of each hemisphere were metallized. This problem is more complicated. You need to solve for the equipotential surfaces and streamlines throughout the solid, subject to the nonconductive boundary condition at the spherical surface.

8. Nov 21, 2012

### haruspex

No, it's quite consistent with what has already been pointed out. Your top cylinder grows narrower tending to 0 area as q tends to ∞. To get a finite resistance you will need to specify a minimum area of contact. Try q cylinders of height (r0-d)/q.

9. Nov 22, 2012

### greswd

Ah that explains it. Thanks haruspex.

Why would there be no conductivity at the surface? Also, if I ignore the above, I should still be able to get an exact Pouillet value?

10. Nov 22, 2012

### marcusl

Last edited: Nov 22, 2012
11. Nov 23, 2012

### greswd

that actually led me to come up with a new problem

12. Dec 13, 2012

### cabraham

Here it is computed, & the answer checks out. Not too hard. If you want to make it hard, offset the terminals at an oblique angle. That is tough. But this sheet computed resistance w/ the terminals diametrically opposite. That is too easy. A single integral solves it. Let me know if clarification is needed. Best regards.

Claude

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13. Dec 14, 2012

### Staff: Mentor

You seem to assume that voltage is a function of the z-coordinate only. Can you prove this assumption? I doubt that it is true.

14. Dec 14, 2012

### cabraham

No, I did not assume that. By symmetry the theta (angle in cylindrical system) vanishes due to z-axis symmetry. The radius "r', does appear in the integral. Take the sphere & divide into flat disk segments centered on the z-axis. The resistance of each disk is ∏c2, where "c" is the radius of the flat disk. I computed said radius on the sheet.

Since the disks are stacked in series, we must sum all disks along the z-axis. The integral is a single integral wrt "z", but the radis of wach disk & that of the sphere enter into the computation. The polar angle theta does not enter in since we have symmetry. If the sphere had a section removed in a way that the angle was less than 2∏ radians, we would have theta in the integral.

Also, if we examine a section where b is half of a, we can compute the resistance of either cylinder with ease. I did that & demonstrated how the sphere section resistance falls in between the inscribed & circumscribed cylinders. That provides a sanity check but does not assure correctness. The resistances of the 2 cylinders are ρ/a multiplied by 0.318 and 0.424. The sphere section resistance must be in between these values. Although 0.350 lies in between 0.318 & 0.424, that itself is no guarantee. There are many values in between 0.318 & 0.424. However if my final answer was outside these values it would be wrong beyond a doubt. Since my answer lies in between these values, it might be correct.

Either way, it cannot be far off. We know the right answer is between the limits I just described. If I erred, please point it out. Did I miss a sign, drop a factor, make an invalid assumption, etc.? Just let me know & I will correct it. BR.

Claude

15. Dec 14, 2012

### marcusl

I have to agree with mfb. The problem with summing the resistances of flat disks is that you are neglecting the lateral spread of current from a small disk at top to a larger disk beneath it. Equipotential surfaces are not normal to the z axis (assuming that electrodes are at z=±a) but rather curve up in the upper hemisphere or down in the lower so as to intersect the spherical surface at right angles. Proper solution of this problem requires solution of Laplace's equation in spherical coordinates, with two sets of BC's specifying a) the source and sink surfaces, and b) zero normal component of current at the spherical boundary. I expect solutions for equipotential surfaces and for streamlines to be expressed in terms of Legendre polynomials.

I don't think this is such a simple problem.

Last edited: Dec 14, 2012
16. Dec 14, 2012

### cabraham

But my math does account for lateral spread of current from smaller to larger disk. I do not believe you read my sheet carefully, but rather skimmed through it briefly. At the top plane, z=b, & there is a surface area, minimum at this plane & its corresponding plane diametrically opposite at z=-b. The area of the disk directly below the top disk is slightly larger than that of the top disk, resulting in lower resistance. The disks are stacked in series so that their resistance values add.

As the cross section changes, the lateral spread of current will encounter a changing area resuslting in changing resistance. I've covered that issue well. Now for the equipotential surfaces not being normal to z-axis, I say they are. Draw a diagram please & attach it for us to review. If you don't, I'll draw one in the evening. Equipotential surfaces intersect the z-axis normally due to symmetry. If, however, the terminal planes were not diametrically opposite, but offset/oblique, you would be correct. That is indeed a much more complex problem.

Please pay attention to my inscribed & circumscribed cylinder resistance values. They illustrate that I did indeed take the lateral current spread into account. Again, I am happy to accept correction, but please review my paper thoroughly. Skimming through it & making off the cuff rebuttals that are meritless gets us nowhere. I'm not out to "win" anything, an argument or otherwise, but I do notice that when someone, not necessarily me, produces info to a tough problem, there are always those who rebuke them while offering no evidence at all where the mistake is. I'm not accusing you of anything, just asking you to support your criticism. BR.

Claude

17. Dec 14, 2012

### marcusl

Claude, your equipotential surfaces are normal to the z axis because you forced them to be so by your choice of solution method. The resistance of a disk, likewise, is only $$R=\frac{\rho l}{A}$$ if the top and bottom surfaces are equipotentials, as though each were coated with a perfect metallic conductor--consistent with equipotentials normal to z. You can see easily that this cannot be a valid solution for the sphere. The BC (that current cannot flow across the sphere surface), together with Ohm's law $$\vec{J}=\sigma\vec{E}$$ means that $$E_{\perp}(r=a)=\left. \frac{\partial\phi}{\partial r}\right|_{r=a}=0.$$ The equipotential surfaces *must* be normal to the spherical surface, not flat as you postulate. It is the curvature of these surfaces that permits current to spread.

Last edited: Dec 14, 2012
18. Dec 14, 2012

### the_emi_guy

19. Dec 14, 2012

### marcusl

The lines should look similar to those for the TE11 mode in a spherical cavity resonator....

20. Dec 14, 2012

### Staff: Mentor

You underestimate the total resistance with your approach as you do not consider current flow in radial direction (which contributes to resistance, too). This is easier to see if you take extreme cases - attach two small disks of radius r (the electrodes) to a cylinder with radius R>>r and height h~r. If you increase R, your calculated resistance would drop with 1/R^2, but the real resistance will approach a finite value.

I did some two-dimensional numerical simulation, but the borders are messy to consider.
For b=0.96, I get ~2.3 ρ/a as resistance, where your formula gives 1.24. I think my simulation overestimates the resistance a bit, so the real value is somewhere in between.
For b=0.52, I get ~0.38 ρ/a as resistance, where your formula gives 0.367. Here, the radial resistance is not significant.

Fixing a=ρ=1 now, as they are just prefactors anyway:

For (b-1)<<1, it is possible to calculate an interesting lower bound of the resistance as following:
Replace the disk (electrode) by a half-sphere. This replaces conducting material by an ideal conductor and clearly reduces the resistance.
Replace the half-ball by a half-ball around the electrode and connect the second electrode to the outer boundary. Use this as half the full resistance. This adds conducting materials where the original problem has none and moves the original symmetry area towards the electrode, so it again underestimates the resistance.
This modified problem has a spherical symmetry, so we can solve it with a one-dimensional integral. Let r be the radius of the electrode, $r^2+b^2=1$.

$$R > 2 \int_r^1 \frac{1}{2\pi r'^2} dr' = \frac{1}{\pi} (\frac{1}{r}-1) = \frac{1}{\pi}(\frac{1}{\sqrt{1-b^2}}-1)$$

Compare this with your formula at b=0.99:
R > 1.93