a solid non conducting sphere of radius R is having charge densityρ=ρ0x where x is distance from the centre of the sphere prove that the self potential energy of the sphere is (πρ0R6)/(6ε0)
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We build up the sphere by adding subsequent infinitesimal layers of charge (carried from infinite distance). From Gauss’s theorem we know that, for an uniformly charged sphere having charge density ρ, radius r, and total charge
q =q(x) = ρ(4πx^3 /3), the field and the potential outside the sphere are those of a point charge q located in the center. Thus, in building the sphere, when a new layer of charge dq = ρ 4πx dx is added, its charge will be located at the potential
V (r) =k q(x)/x, where k = 1/4πε ; thus, the corresponding stored amount of electrostatic energy is V(x)dq. Integrating over r up to the final radius R we find for the total energy U
Uo=∫0RV(x) dq=∫0Rk0q
(x)xρ4πx2 dxUo=∫0R14πε0ρx4πx33ρ4πx2 dx=Uo
q =q(x) = ρ(4πx^3 /3), the field and the potential outside the sphere are those of a point charge q located in the center. Thus, in building the sphere, when a new layer of charge dq = ρ 4πx dx is added, its charge will be located at the potential
V (r) =k q(x)/x, where k = 1/4πε ; thus, the corresponding stored amount of electrostatic energy is V(x)dq. Integrating over r up to the final radius R we find for the total energy U
Uo=∫0RV(x) dq=∫0Rk0q
(x)xρ4πx2 dxUo=∫0R14πε0ρx4πx33ρ4πx2 dx=Uo
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