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 Incoming links: Uniformly polarized sphere

The prescribed surface charge density is that of a uniformly polarized sphere and has zero total charge because

\int cos θ d Ω = 0

t = 0

p_{0} = \int_{r = R} r σ d A = σ_{0} R^{3} \int r cos θ d Ω.

The transverse components cancel by axisymmetry, while the

z

component is

p_{0, z} = 2 π σ_{0} R^{3} \int_{0}^{π} cos^{2} θ sin θ d θ = \frac{4 π}{3} σ_{0} R^{3} .

Thus

p_{0} = p_{0} z, p_{0} = \frac{4 π}{3} σ_{0} R^{3} .

The angular surface distribution is a pure degree-one spherical harmonic. Hence its static external electric potential is exactly that of this dipole, with no higher electric multipole moments; it is not an assertion that the potential inside the finite sphere equals a singular point-dipole potential.

For right-handed uniform rotation about

+ x

, the rigidly attached dipole is

p (t) = p_{0} (- y sin ω t + z cos ω t), \ddot{p} = - ω^{2} p .

Consequently

∣ \ddot{p} ∣^{2} = p_{0}^{2} ω^{4}

\overline{P} = \frac{μ _{0} p _{0}^{2}}{6 π c} ω^{4} = \frac{8 π μ _{0} σ _{0}^{2} R ^{6}}{27 c} ω^{4} \equiv K ω^{4} .

There is no extra factor of one-half: the squared sine and cosine contributions of the two perpendicular oscillating components sum to one. This is the leading electric-dipole approximation for radiation, requiring

ω R / c ≪ 1

and observation in the radiation zone. The stated permanent surface distribution is interpreted as the actual charge producing the given external dipole; adding another dielectric screening factor would contradict the dipole moment the question asks to establish. The relative permittivity enters the induced-polarization problem in part (d), not this prescribed total dipole moment.