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Find the maxim potential a sphere can be loaded; at normal atmospheric pressure,...

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mihuhifitechno | Student, Grade 10 | eNotes Newbie

Posted June 25, 2009 at 9:59 PM via web

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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.

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neela | High School Teacher | Valedictorian

Posted June 26, 2009 at 2:24 AM (Answer #1)

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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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giorgiana1976 | College Teacher | Valedictorian

Posted June 26, 2009 at 10:46 PM (Answer #2)

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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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