Hello everyone,
I was reading Griffiths (9.4.2) and he -when writing the boundary conditions of the electric field in a conductor- said the statement above,("Griffiths : For ohmic conductors (J =sigmaE) there can be no free surface electric currents since this would require an infinite electric field at the boundary" )which I still can't understand. Does anyone have a reason why the free surface currents are zero and the requirement of an infinite electric field.
I was thinking of the dependence of the skin depth on the conductivity (sigma = infinity => delta equals zero => surface currents) however this doesn't imply an infinite electric field.
Thanks in advance
At boundary we deal with E in components form.I think in the mentioned topic they are talking about perpendicular and parallel component of E. The perpendicular component of E is non-zero while parallel component is zero.
The conductor have free charges at the surface that are static. If we have non-zero parallel component of E then there should be current on the surface that can contradict with basic principles of electrostatics.
Thank you for your answer,
You are right about the decomposition of E into a parallel and perpendicular components, however in the boundary conditions
the free charges are only taken for the perpendicular component of the E field, and the free surface currents are in the parallel component of the B field :
ϵ1E⊥1−ϵ2E⊥2=σf
(1/μ1)||B∥1−(1/μ2)||B∥2=Kf×n
Kf is the free surface currents , which Griffiths says doesn't exist in ohmic conductors.
The condition on the parallel component of E is it is continuous not zero.
Thank you once again.
- Badges
- Science topic
- Similar topics
- Condensed Matter Physics
- Zero