A 32 cm -diameter conducting sphere is charged to 530 V relative to V=0 at r=∞?
In the outer points to a charged sphere the electric field behaves as if all the load were at the center of the sphere, so that the surface potential can be expressed as the potential created by an electrical charge Q at a distance r, equal to the radius of the sphere. The potential of a particle has the following expression:
V = Q/4πԐ0r (1)
Q, is the electric charge.
Ԑ0, is the electric permittivity of vacuum.
r, is the distance
The electric charge Q is equal to the charge density σ, multiplied by the surface of the sphere:
Q = σ*S = σ*(4πr^2)
Substituting into the potential equation, we have:
V = (σ*4πr^2)/4πԐ0r = σ*r/Ԑ0
σ = V*Ԑ0/r = (530)(8.854*10^-12)/(0.16)
σ = 2.93*10^-8 C/m^2
With the charge density, we can calculate the electric charge:
Q = σ*S = σ(4πr^2) = (2.93*10^-8)(12.56)(0.0256)
Q = 9.42*10^-9 C
Applying the equation (1)
r = Q/4πԐ0V = (9.42*10^-9)/(12.56)(8.854*10^-12)(21)
r = 4.03 m
r = 4.03 m, from the center of the sphere.