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1. 
   neela Teacher
   Graduate School

   eNotes Editor

   By Gauss' theorem, the electrical intensity, E due to a charge  q on a spherical conductor of radius d is given by :

   E ={1/(4pi*e0)}(q/d^2)

   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.

   =(4pi*1.00054*d^2)*3 M.V

   =37.7195d^2 M.V.

   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.
   

    

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   Posted by neela on Friday June 26, 2009 at 10:24 AM
   
   

2. 
   giorgiana1976 Teacher
   Doctorate

   eNotes Editor

   Best answer as selected by question asker.

   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:

   Emax=Q/(4*pi*epsilon*R^2)=Eo, Q=4*pi*epsilon*R^2*Eo

   Vmax=Q/(4*pi*epsilon*R)=R*Eo

   If R=0.10m, R*Eo=300KV

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   Posted by giorgiana1976 on Saturday June 27, 2009 at 6:46 AM