# Find the maxim potential a sphere can be loaded; at normal atmospheric pressure, discharge in air is produced at the electric field, E0=3.0MV/m.

giorgiana1976 | Student

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

neela | Student

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.