Re: Calculating resistance

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding the dialog between Ludwik K. & John D.:

> > How to solve this problem? Find the resistance R between
> > two small silver-painted dots separated by a distance L on
> > an infinite sheet of a carbon-impregnated paper. The uniform
> > thickness of the paper, d, and its conductivity, rho, are given.
> > Ignore the contact resistance of silver dots.

This problem is exactly solvable in closed form even when
the dots are not small compared to the distance between them,
and even when the dots are of different sizes.

> OK.  Before we get started, let's inventory what we know.
>
> 1) Grand principle: The same equations have the same solutions.

A true and powerful principle.

> 2) Grand principle: Linearity implies superposition.

Also true.  *But* just because we can superpose solutions to obtain
new solutions it is no guarantee that solutions of multiple partial
problems will be the appropriate solution for a composite problem
made of all the partial features.  Sometimes the each of individual
solutions to be superposed do not respect the boundary conditions of
the other individual superposed solution(s).

> 3) Charge is conserved.  Charge obeys a _local_ conservation
> law.  Charge itself is the conserved quantity;  the corresponding
> conservative flow is the ordinarly electrical current.

A useful observation.

> 4) Grand principle:  symmetry.  The situation is
> symmetric under interchange of the dots (and mirror-
> image inversion of the voltage).

Not if the dots are of unequal size.  (But symmetry is still a
useful principle anyway.)

> 5) We seek the steady-state solution.  We inject a steady
> current and wait for transients to die down, then measure
> the voltage.

To do anything else would probably be perverse.

> 6) Electric field is the gradient of the potential.
>
> 7) Ohm's law.
>
> 8) Gauge invariance.  We can choose whatever gauge
> we find convenient.

Which for this problem essentially boils down to choosing a sensible
zero level for the potential.

> > My first idea was to think about the rim-to-rim resistance
> > when the sheet is circular and when one dot is in the center
> > while another "dot" is painted along the circumference.
>
> Basically the right idea.  Let me say it in my own words:
> Ignore one dot.  Analyze the other
> dot surrounded by a circular counter-electrode.  Take
> the limit as the counter-electrode is moved off to infinity.
> Superpose the corresponding solution for the other dot.

There is a problem here.  Unless both dots are infinitely many times
farther apart than their individual radii, the potential from one dot
at the location of the other dot will have its potential change from
place to place over that dot.  This means that the dots can no longer
be thought of as silver with the property that all places on each
dot having a uniform equipotential region if we go with superposing
separate solutions for a single dot where each one is centered at
different places on the sheet.

> > The
> > radius of the central dot is r1 while the radius of the outer
> > "dot" (the radius of the sheet) is r2. This problem is likely
> > to be less complicated and I know how I would start
> > solving it.

When accounting for all current injected at one localized dot by
removing it at the other one we can neglect any outer electrode
at 'infinity' at the edge of the sheet, since no current will
flow asymptotically far from both current carrying dots if the
finite distance between them is held fixed.

> Conservation (item 3) plus stationarity (item 5) plus Ohm's
> law in a homogeneous medium (item 7) guarantees that
> the field is divergence-free.  Equivalently, also use
> the definition of potential (item 6) and conclude that
> the potential obeys Laplace's equation.

Except, of course, on the discontinuity at the boundary edge of the
dots where the silver touches the resistive medium.

> Can't you think of any solution to this equation in
> cylindrical geometry?  Hint:  The same equations have
> the same solutions (item 1).
>
> Additional hint:  Using notions of symmetry (item 4)
> and conservation (item 3) and gauge invariance (item 8)
> we can choose the potential of the counter-electrode
> to be zero.

The problem is not as simple as it looks.  But it *is* still exactly
solvable.

> > This problem has probably been solved somewhere.
>
> Probably.  I think some guy named Fourier had something
> to say about it.
>
> http://www.google.com/search?q=fourier+analytique+chaleur

I didn't bother trying to look up a previous solution, and instead
tried my hand at coming up with the exact solution.  (It was a *whole
lot* easier than finding the exact capacitance of a parallel circular
plate capacitor, which I have yet to make much headway on, BTW.)  The
result is essentially identical to the problem of finding the
capacitance of two infinitely long parallel conducting circular
cylinders, i.e. the capacitance of 'twin-lead' cable problem.

The exact solution for the resistive sheet is problem given below:
Let

R = the net effective resistance between the dots
a = radius of one dot
b = radius of the other dot
L = distance between the centers of the two dots
d = thickness of the sheet/resistive film
[rho] = resistivity of resistive medium in the sheet

To simplify the solution equation and to allow it all to fit on one
line I define two dimensionless auxiliary quantities A & B
defined as:

A == arccosh((L/a + a/L - (b^2)/(L*a))/2) and
B == arccosh((L/b + b/L - (a^2)/(L*b))/2)  .

Then the exact value for R is given as:

R = [rho]*(A + B)/(2*d*[pi]) .

Also, if we want a formula for the capacitance *per unit length* C'
of two parallel conducting cylinders where the distance between
them is greater than the sum of their radii, the exact solution is:

C' = 2*[pi]*[epsilon_0]/(A + B) .

In the special case that both cylinders have the same radius a (= b)
the result for the capacitance per unit length boils down to:

C' = [pi]*[epsilon_0]/arccosh(L/a)

and the formula for the resistance between two *equal size* conducting
dots on an infinite (otherwise uniform) resistive sheet boils down to

R = [rho]*arccosh(L/a)/([pi]*d) .

I don't know if these formulae match those given in the reference
mentioned by Bernard Cleyet or not.  But if these don't they might
still be shown to be equivalent by making appropriate mathematical
manipulations involving inverse hyperbolic function identities.

David Bowman
David_Bowman@georgetowncollege.edu