2 Answers | Add Yours
Outside of a sphere, the potential and the field are the same as the sphere's load would be centered.
V=Q/(4*pi*epsilon*r) and E=Q/(4*pi*epsilon*r^2), r>R
At the sphere's surface, the field is maxim, so appears the risk of abrupt discharge:
If R=0.10m, R*Eo=300KV
By Gauss' theorem, the electrical intensity, E due to a charge q on a spherical conductor of radius d is given by :
Where e0 is a dielectrical constant for air which is 1.00054.
The discharge intensity is given to be E0 =3.0 MV/m/
Therfore,the maximum potential q that can be loaded on the spherical conductor is given by:
q = 4pi*e0*d^2*E0.
Therefore the maximum potential that can reside on a spherical conductor of radius d meter is 37.7195*d^2 M*V/m
Hope this helps.
We’ve answered 333,958 questions. We can answer yours, too.Ask a question