Solution

P

is electric dipole moment per unit volume. It produces

ρ_{b} = - \nabla \cdot P, σ_{b} = P \cdot n .

D = ϵ_{0} E + P

is useful because

\nabla \cdot D = ρ_{free},

so it separates free charge from the bound charge represented by

P

. In a linear isotropic dielectric,

D = ϵ E

Spherical symmetry and Gauss's law for

D

give, on both sides of the interface,

D = \frac{q}{4 π r ^{2}} e_{r} .

Thus

P_{i} = (1 - \frac{ϵ _{0}}{ϵ _{i}}) \frac{q}{4 π r ^{2}} e_{r} .

With

n = e_{r}

pointing from medium one to medium two, the net bound sheet charge at

r = R

is the polarization jump

σ_{b} = (P_{1} - P_{2}) \cdot n = \frac{ϵ _{0} q}{4 π R ^{2}} (\frac{1}{ϵ _{2}} - \frac{1}{ϵ _{1}}) .

If one assigns surface charge separately to the inner body and outer medium, their contributions are

P_{1} \cdot n

and

- P_{2} \cdot n

; their sum is the displayed physical interface charge.

Solved by gpt-5.6-sol high.

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