Consider a semi-conductor in which conduction is mainly due to free electrons(valence electrons that are free to roam throughout the material) , and since it's a semi-conductor say not all atoms contribute these free electrons(some do,some don't ) and under the application of an external electric field these free electrons start moving in the material and the other non-contributing atoms(atoms that do not contribute any electrons for conduction) get stretched as in the case of a dielectric .I have the following questions :

1 )The neutral non-contributing atoms get stretched as in the case of a pure dielectric,but do the contributing ionised atoms also stretch a bit? I do think they stretch but i wonder if they contribute anything to the polarization vector $\vec P$ .

2 ) In calculating $\vec P$ should we also consider the free conduction electrons ,because they are also from the atoms of the material and we want the total bound charge to be zero ,i.e when we write Div.$\vec P =-\rho_b$ , is the charge due to these conduction electrons included in $\rho_b$? I think it should be included because we wish that the continuity equation hold for these bound charges and that requires total bound charge to be constant,if these conduction charges are not included then continuity equation won't hold since an atom which is non-contributing might release an electron if the electric field let's say is increased ,then this electron now no longer is bound,hence for this reason i think all the charges belonging to the material be counted as bound.

3)If the charge due to conduction electrons is also included in $\rho_b$,then is the conduction current also somehow included in the polarization current $\frac{d\vec P}{dt}$?


1 Answer 1


I have not worked with semiconductors, but will take their properties as you describe them.

  1. The ionized atoms should contribute, but their polarization will be different.
  2. Conduction electrons are free to move, forming a current, and will not be included in P.
  3. Conduction electrons are never included in the polarization.
  • $\begingroup$ If conduction electrons are not taken as a part of the bound charges, can we still say that the continuity equation will hold for the bound charges? I'm asking this because if the electric field is changed (say increased) more atoms could lose electrons and these electrons which once were bound are no longer to be considered bound $\endgroup$
    – Arjun
    Oct 6, 2023 at 7:53
  • 1
    $\begingroup$ Conduction electrons can not taken as a part of the bound charge. The continuity equation applies to free charge and current, and for bound charge and current, separately $\endgroup$ Oct 6, 2023 at 18:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.