Understanding Ohm's Law and Charge Units in Electrodynamics

[yungman]

This is referring to Chapter 7.1, page 285 of "Introduction of Electrodynamics" 3rd edition by David Griffiths.

$$\vec J = \sigma \vec f \;\; \;\;\;\;\;\;\;( \frac{C}{m^2\cdot sec})$$ (1)

Where $f$ is force per unit charge. Is the unit charge one coulomb?.

Also

$$\vec J = \sigma(\vec E +\vec v X \vec B)\;\;\;\;\;\;\;\;\;\;\; (\frac{C}{m^2\cdot sec} )$$ (2)



My understanding is force:

$$\vec F = q(\vec E +\vec v X \vec B) \;\;\;\;\;\;\;\;\;\; \hbox {( N) }$$

For unit charge of one coulomb,

$$\vec f = (\vec E +\vec v X \vec B)$$ (3)


How do I go from (2) to (3)

Thanks

 

[LostConjugate]

yup, the unit charge is a coulomb

Equation 1 shows how you get from 2 to 3

 

[yungman]

yup, the unit charge is a coulomb

Equation 1 shows how you get from 2 to 3

Thanks for the reply.

So if $\vec f$ is unit force which is the force of a coulomb charge.

$$\vec F = Q\vec E = 1\;X \;\vec E = \vec E \hbox { for Q = one coulomb }$$?

Therefore

$$\vec J \;=\; \sigma \vec E = \sigma \vec f$$

if the velocity of the unit charge is slow and

$$Q (\vec v X \vec B)$$ is ignored.

Am I getting this right? Just that simple? What was I thinking!

Thanks

Alan