Re: [Phys-L] amusing electrostatics exercise

[Bruce Sherwood <Bruce_Sherwood@ncsu.edu>]

Whether my current distribution is realizable or not is irrelevant to using
it to calculate what the field would be, since the fields will be the same
as in the stated situation. You don't have to have physical charges for the
use of mirror charges in electrostatics to be valid. However, here's a way
to do it:

Fill the hole with a large number of very thin wires. Half of them have
current running to the left, half to the right, whereas the outer material
can be solid and has current running to the right. The currents of course
have to be chosen to be consistent with that in the outer metal. With fine
enough wires, this is the functional equivalent of my distribution.

I don't understand the charge conservation issue. If there is an issue,
there's a problem with a single very long current-carrying wire, a problem
we typically sweep under the rug (or we consider a coax cable).

Bruce


On Wed, Feb 27, 2013 at 11:23 AM, John Denker <jsd@av8n.com> wrote:

> On 02/27/2013 08:14 AM, Bruce Sherwood wrote:
> > One way to handle this is to consider the superposition of the field of a
> > solid wire and the field of a wire the size of the hole, with current
> > running in the opposite direction. Ampere's law gives you the field at
> any
> > location for each of these current distributions; add them vectorially.
>
> How do you do that in a way that upholds conservation of charge?
> As another way of saying the same thing, how do you make the current
> in the hole become part of a complete circuit?
>
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