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Re: I need help. (long)



Regarding Ludwik's claim that an accurate value for the overall
interelectrode resistance assures accurate equipotential curves (when
the experiment is done on finite paper and the theory is for an
infinite sheet):

But what is wrong with my claim that the finite size of the sheet
has a negligible effect for the geometry chosen?

It probably *does* have a negligible effect on the overall value for
the interelectrode resistance. But this does not mean that it has a
negligible effect on the potential values out in the parts of the
paper that are distant from the close vicinity of the electrodes, and
it does not mean that the equipotential curves out there are only
negligibly distorted from circles.

Is it not true that
rho calculated from your formula, David, and rho calculated from
the narrow strip would be different if a sizable fraction of the total
flux was prevented from going through regions outside the paper?

Not necessarily.

The two values of rho turned out to be practically identical. How
can you ignore this argument?

I'm not ignoring it. I'm discounting it.

I will ask students to locate the experimental curves for a geometry
in which circles are closer than 10 cm. If the discrepancy between
what is observed and what is predicted is reduced then I will also
start blaming the paper size effect. We will see. For the time being
I will blame surface charges. I know that such blaming has no value
unless some theoretical or experimental evidence is produced. I think
I do have some evidence against the paper size effect.

You *do* seem to have some evidence against a significant finite
size effect in the value of the overall resistance. But you do not
have such evidence for the potential values themselves out in the
far field.

Where am I wrong with it?

I think your conceptual problem is that you think an accurate value
for the overall resistance ought to imply a corresponding relative
accuracy in the values of the potential itself. This is not the
case. It is quite possible to have a reasonably accurate value for
the resistance (or for the capacitance for the corresponding
capacitor problem) and still have reasonably inaccurate values for
the potentials themselves.

'Why is this?', you ask. Well, it has to do with the way the overall
resistance (and/or capacitance) is calculated in terms of the
potential field used. It ends up that the overall capacitance/
(resistance) is related to the value of the electrostatic energy in
the field for a given potential field configuration. A variational
principle is at work here. The energy stored in the field is
*minimized* (over the set of all possible potential fields
conceivable that obey the specified boundary conditions) by the
particular correct potential field function that properly satisfies
Laplace's equation throughout the region of interest. Suppose we try
to evaluate the energy in the field using an incorrect potential
function. Then the stationarity of the energy over the family of
possible potential functions implies that first order errors in the
value of the potential translate to 2nd order errors in the value of
the field energy calculated with that errant potential. Suppose the
potential is in some sense off from the correct value by about 10%.
Then we expect that the field energy (and hence the capacitance and
the paper's resistance) are all off by something like 1% from their
correct value. Effectively, the error in the resistance (& field
energy, & capacitance) is proportional to the *square* of the error
in the potential used to calculate it.

If you remember how the variational method works for finding
approximate ground state energy eigenvalues in quantum mechanics, you
can see that the same underlying mathematical effect is at work here.
Suppose you want to calculate an approximate (but reasonably accurate)
value for the ground state energy for some Hamiltonian that is too
difficult to exactly solve for. What you do is make up a 'trial wave
function' that is supposed to reasonably well represent the actual
ground state wave function in that it is qualitatively correct, but
quantitatively off. This 'trial wave function' typically has one or
more adjustable parameters in it that can be tweaked to improve the
agreement between the actual ground state eigenfunction and the
'trial wave function'. What you then do is evaluate the expectation
of the Hamiltonian using the 'trial wave function' for generic values
of the adjustable parameter(s). You then minimize this expectation
over the set of possible values for the parameter(s). This lowest
possible energy expectation over the family of 'trial wave functions'
is usually quite close to the actual ground state energy value (but
is still a wee bit high) for the Hamiltonian in question. The
value(s) of the adjustable parameter(s) that minimize this energy
expectation defines the so called 'best possible' 'trial wave
function' to represent the actual ground state given only the choices
presented being the members of the family of 'trial wave functions'.
This 'best' wave function is typically *still* significantly quite in
error relative to the true ground state wave function. But the first
order errors in this wave function show up as only second order
errors in the calculated value for the ground state energy. Thus the
method works surprisingly well for finding ground state energy value
given the crudeness of the family of 'trial wave functions' that one
typically has to work with for reasons of tractability.

In fact, you can look at the expression of the electrostatic field
energy *as* a hidden form of a quantum mechanical energy expectation.
The field energy is proportional to the spatial integral of the
square of the electric field (or equivalently, to the square of the
gradient of the potential). If we require that this expression is
minimized w.r.t. all possible potential functions that obey the
specified boundary conditions, then the one that minimizes it *is*
the correct solution. Also a little use of the calculus of
variations shows that the condition of minimization of the
electrostatic energy is equivalent to the requirement that the
potential field satisfy Laplace's equation.

Now if we take the electrostatic energy (in terms of the square of
the gradient of the potential) and integrate it by parts, we can
convert it into an expression in which is proportional to the
'expectation' of the negative of the Laplacian operator sandwiched
between two factors of the potential field. In our electrostatics
problem the potential plays the role of the wave function. The
field energy plays the role of the ground state energy value and the
negative of the Laplacian operator plays the role of the
Hamiltonian operator in the quantum mechanical analog. In fact,
since the QM kinetic energy operator is also proportional to the
negative of the Laplacian operator, we see that the QM problem of
finding the ground state for a(n otherwise) free particle in a box
with some specified boundary conditions on the box walls can be
thought of in terms of a corresponding electrostatics problem.
Because of this analogy, we can use a corresponding variational
method to solve electrostatics problems that is analogous to that
used in solving for QM ground state problems. The main difference
is that the constraints needed for the 'trial wave function' are
typically different. In the electrostatics case the 'trial
potential function' needs to appropriately obey the given boundary
conditions, and it doesn't have the QM normalizability constraint
that the integral of its square is 1.

There is another variational method that can be used in solving
electrostatics problems that may be somewhat more useful in
finding capacitances than the one mentioned above. This other method
doesn't try to find the potential field function everywhere, but
instead assumes a particular 'trial surface charge function' on the
relevant conducting boundary surfaces. This surface charge function
is then adjusted to minimize the total electrostatic energy that it
produces via the coulomb potential between all pairs of charges.
The nice thing about this method is that it is a lower dimensional
problem than the other method. (I.e. the surface charge
distribution is defined on a lower dimensional domain than the
potential function is.)

The reason that the capacitance C for a given 1 or 2 conductor
electrostatics problem is related to the electrostatic field energy
is simply because that field energy satisfies U = (C/2)*V^2 where the
V is either the potential on a single conductor if the potential at
infinity is zero, or V is the difference in potential between the 2
conductors in the 2 conductor problem. Once U is calculated for a
given V (specified by the boundary conditions) we immediately have C.
And in the case of the 2-d conducting sheet the overall resistance
R is related to the corresponding capacitance per unit length C/l via
the replacement C/(l*[epsilon_0]) --> 1/R and the integral expression
for the total field energy in the electrostatics problem becomes a
corresponding integral expression for the total power dissipation
rate in the DC conduction case (recall then the power dissipated is
V^2/R). The DC conduction problem is solved by whatever potential
field configuration that minimizes the overall dissipation rate for
a given potential difference between the electrodes.

David Bowman
David_Bowman@georgetowncollege.edu