A sphere with radius R=10 cm has the potential V=10KV,in the air. Which is the load superficial density, on the sphere?

giorgiana1976 | Student

The sphere's potential is

V=Q/(4*pi*epsilon*R), so Q=4*pi*epsilon*R*V

(epsilon is electrical allowance of the environment)

sigma=Q/S=Q/(4*pi*R^2)=(4*pi*epsilon*R*V)/(4*pi*R^2)

sigma=epsilon*V/R=0.88microC/m^2

neela | Student

By Gauss' Theorem, the entire charge is distributed over the surface of the sphere.

The curvature of the sphere is unoform over the entire surface.So, the load is also spread uniformly over the entire spherical surface, 4pi*r^2, where  r  is  the radius of the sphere.

The  density ,E of the charge is the charge  per unit area  and is given by:

E =( q/(4pi*r^2))/e_0                         (1) and

q= 4pi*e_0*r*V ,  e_0 being permittivity of air and is nearly 1,                      (2).

From one and (2):

E= V/r =

Given r= 10cm = 0.01 meter and

V=10KV= 10000Volts = 10^4 volts. Substitute in (1) to get:

Superficial density of load, E = 10^4/(0.01) =10^6 volt/m