36D (Electrodynamics)

Insert

ρ = ϵ_{0} \nabla \cdot E

and

J = μ_{0}^{- 1} \nabla \times B - ϵ_{0} E_{t}

into the Lorentz force and use vector identities plus Faraday's law. Rearrangement gives

\partial_{t} g_{i} + \partial_{j} σ_{ij} = - f_{i}

. Here

g = ϵ_{0} E \times B

σ

Solved by gpt-5.6-sol high.

b

With no free surface charge, tangential

E

and normal

D

are continuous. The image ansatz gives

\frac{q ^{'}}{q} = \frac{1 - ϵ _{r}}{1 + ϵ _{r}}, \frac{q ^{''}}{q} = \frac{2 ϵ _{r}}{1 + ϵ _{r}} .

z = 0^{-}

, with

R = (r^{2} + d^{2})^{1/2}

E_{r} = \frac{2 q}{1 + ϵ _{r}} \frac{r}{4 π ϵ _{0} R ^{3}}, E_{z} = \frac{2 ϵ _{r} q}{1 + ϵ _{r}} \frac{d}{4 π ϵ _{0} R ^{3}} .

Solved by gpt-5.6-sol high.

Integrating the normal Maxwell traction (the radial parts cancel) gives

F_{z} = \frac{q ^{2}}{16 π ϵ _{0} d ^{2}} \frac{ϵ _{r} - 1}{ϵ _{r} + 1} .

It points toward the dielectric when

ϵ_{r} > 1

. This equals the Coulomb force from the image charge, as it must.

Solved by gpt-5.6-sol high.

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