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# smooth sphere vs unsmooth sphere

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From:  Antonio C. M. de Queiroz [SMTP:acmq-at-compuland-dot-com.br]
Sent:  Saturday, April 04, 1998 12:48 AM
To:  Tesla List
Subject:  Re: smooth sphere vs unsmooth sphere

Sulaiman Abdullah wrote:

> Yes, this rersult is near enough to a similar 'rule of thumb' which
> gives 3 kV / mm whether in a 'gap' or at the edge of a sphere.
>  V = 2.83 Mv x R  (R = radius of sphere in meters) ... comments?

The electric field at the surface of a sphere with radius r charged
with a charge q is:
E=q/(4*pi*e0*r^2) (e0=8.85e-12, r in meters, q in Coulombs).
The voltage at the surface of the same sphere is:
V=q/(4*pi*e0*r)
And so:
V=r*E
If the maximum E without air dielectric breakdown is ~3 MV/m for DC:
Vmax=3e6*radius.
Maybe somewhat less for high frequency.

Note that for the same Emax, the surface charge density in the sphere
is:
qmax/area = qmax/(4*pi*r^2) = e0*Emax =8.85e-12*3e6=26.6 uC/m^2.
If you use 25 uC/m^2 instead, Emax=25e-6/8.85e-12=2.82e6,
and the rule for Vmax become Vmax=2.82 MV*r.

Antonio Carlos M. de Queiroz
http://www.coe.ufrj.br/~acmq

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