Question

Griffiths Intro to Electrodynamics, 4th ed., 7.1:

Two metal concentric shells, with inner radius a and outer radius b, are separated by a weakly conductive material with conductivity σ. The shells are held at potential V. The question asks for the current that flows from a to b. My question is not how to do this, but: why do I not need to take into account the permittivity of the material, and hence the polarization of it?

My problem solving process goes as follows:

V = -∫E⋅dl = IR

∮E⋅da = Q_enc/ε₀ Why can I directly find E and use ε₀, as opposed to using D (electric displacement) and the relative permittivity? I get that the material between the shells is still conductive, despite it being weak, but isn’t any given dielectric?

E = (Q_enc) / (4πr²ε₀)

V = -[(Q_enc) / 4πε₀] ∫(1/r²)dr, from a to b

= [(Q_enc) / 4πε₀][(1/b) - (1/a)]

Q_enc = (4πε₀V) / [(1/b) - (1/a)], so

E = { (4πε₀V) / [(1/b) - (1/a)] } / (4πr²ε₀)

= V / { r² [(1/b)-(1/a)] }

J = σE

= σV / {r²[(1/b)-(1/a)]}

I = ∮J⋅da

= ∮(σE)⋅da

= { σV / ( r² [(1/b) - (1/a)] )} ∮r²sinθdθdφ

= 4πσV / [(1/b) - (1/a)], in r direction.

Answer 1

You should know that electric field inside the medium is always zero and charge is always reside on its surface i.e. there is no charge inside it. But there constant potential V =Va -Vb. We have to find E but for that we have to know the charge Q on the inner shell that's why in the calculation Q has founded in the form of V .But Q-enc in the calculation doesn't mean charge enclosed in the conductive medium it's the charge on the inner shell .Since we have calculated for the charge on the inner shell we don't require permittivity of the conductive medium.