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# Re: electromagnetic shielding

• From: Leigh Palmer <palmer@SFU.CA>
• Date: Thu, 1 Feb 2001 09:56:39 -0800

A related question:

Is it a simple matter to calculate the electric field in the region
between a circular cylinder of charge Q1 and an elliptical cylinder
of charge Q2? Assume the circular cylinder has a radius a < b (the
semi-minor axis of the elliptical cylinder), they share an axis of
symmetry, and the region in between is a vacuum.

The outer cylinder must be nonconducting to meet your specification.
If it is a conductor, Q2 = -Q1. (I assume the cylinders are long.)
You should also specify the length and semimajor axis, and specify a
constraint on the charge distribution (e.g. uniform surface charge
density on the elliptical shell). Simple versions of this problem are
solved simply using the principle of superposition: the field due to
the two cylinders is just the (vector) sum of the fields due to the
individual cylinders. They don't interact.

Since your subject line refers to shielding I will assume that you are
thinking of the case in which both cylinders are conducting.

I know that if the outer cylinder is also circular the electric
field in between is a function of Q1 and r only, but I suppose that
if the outer cylinder is not circular this is no longer true. Is
this correct?

Justin Parke
Oakland Mills High School

Dear Justin,

You are correct; it is no longer true. I'm very rusty on what problems
are capable of solution in closed form, but if the one you propose is
one of those, it is too hard for me to solve, and the mathematics may
be over your head, too. Failing a closed form solution, let me suggest
a couple of approaches you might follow.

Lines of force (or electric field) must intersect the surface of a
conductor normally (perpendicular to the surface) if no currents are
flowing. The number of lines of force per unit area is proportional to
the electric field strength. The electric field is strongest along the
semiminor axis of the ellipse. You should draw what you think the
field might look like if it had to conform to these conditions. That

Although the mathematics of this problem is difficult, it is amenable
to approximate solution by numerical means. A relaxation calculation
is easy and fun using a spreadsheet (like Excel) and one of the
improbably powerful computers that seem to be so common today. In this
case the geometry is difficult, but not impossible. You would be well