I am using the approximation that this cylinder is very long, longer than
the other dimensions, then I can solve it as a two dimensional problem and
find the capacitance per unit length. I am also assuming that the plane is
much larger than the distance D to the cylinder of radius R, then I can
approach the problem  as  PEC cylinder  above a PEC plane, an image plane.
then by the image theory you can substitute the flat plane by a cylinder of
the same dimensions as the other one but with different polarity. So the
equivalent problem is now two cylinders of radius R separated by 2*D. that
is an analytical problem that can be found in the literature (among them
Cheng "fundamentals of engineering Electromagnetics" Addison Wesley) and
whose solution I posted in a previous message 

Vicente Rodríguez, Ph.D., E.I.T.
RF/Electromagnetics Engineer
ETS-Lindgren
(an ESCO Company)
P.O.Box 80589, Austin TX 78708-0589
phone 512.835.4684 x648
fax 512.835.4729     
[email protected]
http://www.emctest.com
http://home.austintx.com/~vicenter

> -----Original Message-----
> From: Cortland Richmond [SMTP:[email protected]]
> Sent: Monday, December 18, 2000 10:56 AM
> To:   Vince Rodriguez
> Subject:      RE: Capacitance calculation
> 
> Vince Rodriguez <[email protected]> wrote:
>  
> 
> >>You can use image theory and just get the capacitance between two
> cylinders separated by 2D<<
> 
> That doesn't sound right. How is the capacitance between two objects
> separated by some distance going to be equalled by that between two
> objects
> (and one of them is smaller than before) separated by TWICE the distance? 
> 
> This is an experiment that can easily be done with common test equipment.
> 
> Cortland

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