 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2b/section-b/b6/solution

D = σ_{2} d_{1} + σ_{1} d_{2}

. Then

E_{1} = σ_{2} V / D

E_{2} = σ_{1} V / D

J = σ_{1} σ_{2} V / D

. The interfacial free and total charges are

(ϵ_{2} σ_{1} - ϵ_{1} σ_{2}) V / D

and

ϵ_{0} (σ_{1} - σ_{2}) V / D

, respectively.

Full derivation: solution.

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