Re: I need help.

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

Regarding Ludwik's question:

> ...    I wander what you think about
> my "E line bulging" questions? They are asked under
> the above URL. The basic question is:
>
> "Why does the flux of E lines prefer a thin conductive
> sheet (or why does it avoid the surrounding air)?"
>
> I am referring to the geometry in which two small silver
> dots are separated by 10 cm. To avoid complications
> assume that the paper size is practically unlimited.
> Ohm's law could be used to answer a similar question
> about the flux of current density lines but it can not be
> used to explain the behavior of E lines. What controls
> the behavior of electric field lines in our 3D space?  ...

Ludwik, a an appreciation of a few facts is necessary to understand
why the conducting paper system acts like a 2-d electrostatics
problem.  First, the electric field is the negative of the gradient
of the potential both on the paper *and* off of it.  Second, we have
Kirchoff's current law applying to any and all steady state currents
flowing. And third we have the current density locally related to the
electric field by ohm's law.  These facts are sufficient to determine
why the field pattern in the paper follows that of an electrostatic
problem a 2-d world, and *not* a 3-d world.

Fact one tells us that E = - grad(V).  A continuous version of fact
two tells us that div(j) = 0.  Fact three tells us that
j = [sigma]*E. Here [sigma] is the conductivity tensor for our
medium and the * product is a contraction product between the
second rank conductivity tensor and the vector E.  The result of
this tensor product (effectively equivalent to the product of
a matrix times a column vector being another column vector) is the
vector j.  Putting all three of these equations together gives:

0 = div([sigma]*grad(V)) .  (Here we cancelled out the - sign from
the E field being the negative gradient of V.)

Now if [sigma] is a bulk isotropic uniform conductivity in all
directions then its tensor is a scalar constant times the 2nd rank
identity tensor and we can factor that constant out of the
divergence and then divide both sides of the equation by that
constant.  The result in that case is 0 = div(grad(V)) (i.e.
Laplace's equation in 3 dimensions).

But for the Pasco paper case the conductivity tensor is manifestly
not uniform in space nor is it isotropic (when we consider the paper
as effectively infinitely thin and we look at the conductivity in the
paper's plane).  For this paper case the conductivity tensor is
manifestly the zero tensor everywhere outside the paper since air
is an electrical insulator.  This means that our above equation
boils down to the trivial identity 0 = 0 everywhere outside the
paper.  But *on* the paper the conductivity tensor is nonzero.  On
the paper the conductivity is isotropic in the x-y plane, but it is
essentially zero along the z direction  perpendicular to this
conductive plane.  This is because no current is capable of jumping
into or out of the conductive plane along the z direction.  This
means that for locations *on* this conductive plane the
conductivity tensor has the matrix structure:
                     | 1 0 0 |
[sigma] = (constant)*| 0 1 0 |
                     | 0 0 0 |

The fact that the zz diagonal component of this tensor vanishes
means that this tensor acts as a projection operator that
projects the E-field into the x-y plane and eliminates its
z-component.  We know this must be the case because the current
vector must be confined to the conductive plane.  Since the
z-component is missing from the grad(V) after it is multiplied by
[sigma] this means that when the subsequent divergence is taken
that divergence has no contribution from the z-component of
grad(V).  The result is the *2-dimensional* Laplace's equation:

0 = div(grad(V)) (2 dim) = d^2(V)/dx^2 + d^2(V)/dy^2 = 0

where the derivatives are supposed to be understood as partial
derivatives.  Thus inside the conductive plane we have
the potential function obeying the 2-d Laplace's equation
(giving 2-d behavior for the potential, electric field, and
current in the paper), and outside the conductive plane we have
the indeterminate degenerate identity  0 = 0.

If we are only interested in what's going on in the conductive plane
we can now understand why it behaves completely like a 2-d
electrostatics problem, and why the E-field outside that plane is
totally irrelevant for that behavior.

David Bowman
David_Bowman@georgetowncollege.edu