Poll Results
No votes. Be the first one to vote.
Participate in Poll, Choose Your Answer.
Calculate the electric field when the conductivity is 20 units, electron density is 2.4 units and the velocity is 10m/s. Assume the conduction and convection current densities are same
The electric field (E) in a conductor, where the conduction and convection current densities are considered equal, can be described in terms of the conductivity ((sigma)), electron density ((n)), and electron drift velocity ((v_d)) using Ohm’s Law and the definition of current density.
Ohm’s Law in terms of current density ((J)) and electric field ((E)) is given as:
[J = sigma E]
The current density can also be related to the electron charge ((e)), electron density ((n)), and electron drift velocity ((v_d)) as:
[J = n e v_d]
Given that the conduction and convection current densities are the same, we set these two expressions for (J) equal to each other:
[n e v_d = sigma E]
Rearranging this equation to solve for (E) gives:
[E = frac{n e v_d}{sigma}]
However, to compute the value of (E), we need the value of the electron charge ((e)). The charge of an electron is a constant, approximately equal to (1.6 times 10^{-19} , C) (Coulombs).
Given:
– Conductivity, (sigma = 20 , text{units}) (assuming SI units, (text{S/m}))
–