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Re: Charged disk; was electrostatic ...



If there is nothing special about the geometry, why not model it as two
spheres that are very far apart connected by a thin wire (so that the
potentials can be approximated by kQ/r)?

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| Robert Cohen Department of Physics |
| East Stroudsburg University |
| bbq@esu.edu East Stroudsburg, PA 18301 |
| http://www.esu.edu/~bbq/ (570) 422-3428 |
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On Mon, 5 Feb 2001, Ludwik Kowalski wrote:

My guess (again only a guess) would be "close to 16, not
close to 4". Why? Because the sphere whose radius is 4
times larger has 16 times larger area. But I do agree that
a numerical calculation would be much better than a guess.
I do not know how to solve this problem; if I did I would
write a "quick and dirty program" to solve it numerically.
I hope somebody will do this for us.
Ludwik Kowalski

Leigh Palmer wrote:

Can somebody answer this question? Suppose the electrified
object is like a pear. Or more exactly it is a set of two spherical
segments connected smoothly with a truncated cone. Let the
left segment have R1=20 cm, the right segment R2=5 cm and
the distance between the centers of segments 30 cm. There is
nothing special about this geometry; it is an illustration.

We all known that concentration of charges will be higher on
the right "pole" than on the left "pole". But how many times
larger? Factor of 4? Factor of 16? Neither of the above? Is
there any rule of thumb about this?

The answer will be nearer to 4 than to 16, but it is also well
known that if you wish to solve this problem you must do so
numerically.

Leigh