Natural Science Forum / Physics / Electromagnetism / July 2009

In article
<fc5d36ce-19c7-4d37-9048-3d145dbefb6f@g1g2000yqh.googlegroups.com>,

There is an example given in Ramo and Whinnery's "Fields and Waves in
Modern Radio." The same techniqu can be extended to cover this case.

The potential for a sphere in the presence of other conductors is going
to be a sum of spherical harmonics of unknown coefficients. This sum has
to equal the potential on the sphere and the cylindrical  wire. There
should be sufficient symmetry to set some of the harmonics to zero
amplitude. Matching potentials at 1000 points will give a linear
equation set. Even modest computers these days can handle the solution
of such equations.

I am not saying that it would be easy but it is certainly feasible.

Bill

Hi, Wim,

We have two fields
(a) a field E=rho/(2*pi*eps0)*r) due to the line (radial with respect to the
line) and (b) a field E=q/(4*pi*eps0**r^2) due to the conductor charge. rho
is the charge per unit length of the line and q is the charge on the sphere.
Assuming the theoretical infinite line on which the line charge  is based
and assuming that the line/ sphere system is "isolated" then it appears that
the sphere's charge will be rather large so that its field will dominate in
the near region and your expression gives a valid limit based on completely
ignoring the field due to the line charge (assuming that the line has some
physical radius).

On that basis, I agree with you.

However, power "lines" invariably have  more than one conductor (and or
images)and also do involve a ground plane and images so I also have to agree
with your caveat.
If, as indicated, the sphere is small and its net charge is near 0, it will
not disturb the field due to the line(s). Then only the first field above
need  be considered- naturally this will  give a quite different result- but
one than may well be more realistic.   An approach using potential
coefficients can be used as for a conductor instead of a sphere and either
get the open circuit voltage for a given line charge or a short circuit
current for sphere at ground potential  (in this case it is per unit length
so an approximation would be a conductor of length twice the sphere radius)
. This, of course is an approximation that will give a different answer than
your approximation based on the original question so all that I can suggest
is something in between or the approach that Salmon Egg suggests.

Don
dhky@shawcross.ca
remove the x to reply

On 5 jul, 02:16, " Don Kelly" <d...@shaw.ca> wrote:

Hello Don,

You right, there is no radial symmetry.  When assuming: power line
very large with respect to the sphere's diameter and distance between
power line large with respect to sphere's diameter, most of int(E*ds)
is close to the sphere.

Close to the sphere there is reasonable radial symmetry. So as a first
estimate, you can use the capacitance for a sphere, where diameter of
outer sphere goes to infinity.

When requiring better accuracy, you need to do the math, and also
taking into account mutual capacitances via other structures.

Best regards,

Wim

Wimpie nearly gets the prize. There is a lot of w.nking going on here,
but a real scientist/engineer/physicist knows how to THINK about the
problem and home in on some reasonable approximations. In this case
there is a sphere and a power line forming a capacitor. If we think
about this we can see that this situation is sort of half way between
the self-capacitance (the sphere and infinity) and a spherical
capacitor where the inner radius is .05m and the outer radius is 1
meter. [in essence the power line is expanded and stretched to form a
terminal all about the inner sphere]. These represent the lower and
upper limits of the capacitance of the capacitor in question.
Interestingly, if you calculate this you find the self-capacitance is
about 5.56 pf and that of the spherical capacitor is roughly 5.86 pf.
Hence the capacitance of the system in question will be somewhere
between those two values. I'm guessing the true value will be closer
to the low value than the high one.

Exactly. If you really need extreme accuracy, everything in the area
must be specified and the actually capacitance calculated either
numerically with a computer or if possible by some theoretical method
(which probably won't take into account other structures).