# Why should the relation (current density)= (resistivity)×(velocity of charge) [$j = ρ v$] be applied to each type of charge carriers separately?

This is from the class 12 physics NCERT Part I.

The relation $$\mathbf{j} = \rho\mathbf{v}$$ should be applied to each type of charge carriers eeparately. In a conducting wire, the total current and charge density arises from both positive and negative charges:
$$\mathbf{j} = \rho_+\mathbf{v}_+ + \rho_-\mathbf{v}_-$$
$$\rho = \rho_+ + \rho_-$$
Now in a neutral wire carrying electric current,
$$\rho_+ = -\rho_-$$
Further, $$v_+ \approx 0$$ which gives
$$\rho = 0$$
$$\mathbf{j} = \rho_-\mathbf{v}$$
Thus, the relation $$\mathbf{j} = \rho\mathbf{v}$$ does not apply to the total current charge density.

I'm having trouble understanding this point, mainly the part about the positive charge velocity being zero and the need to apply the formula separately for positive and negative charges. Any help is appreciated.

• the part about the positive potential being zero? Note that $\mathbf{v}$ means "velocity", but not "potential". – Thomas Fritsch Jul 13 at 11:19
• oh you're right! thanks!! – laksheya Jul 13 at 12:25

1. The fact that the wire is electrically neutral means, that the positive and negative charge densities are equal. Hence, $$\rho_+ = \rho_-$$. Since the the total charge density is simply the sum of these two densities we obtain $$\rho_{tot} = \rho_+ + \rho_- = 0$$. This simply states, that the wire is electrically neutral.
2. The fact that the protons are bound in the solid structure of the wire means, that they can not change their position. Hence, their mean velocity vanishes, $$v_+ = 0$$. In contrast the electron gas is free to travel along the wire. Thus if we apply a voltage across the wire, they move towards the anode (=positively charges side). Hence, their mean velocity is non-vanishing, $$v_- \ne 0$$.
3a. In principle a current density (or simply a current) can be due to a either a movement of the positiv or the negativ charges in the body. Thus, $$j_{tot} = j_+ + j_- = \rho_+ v_+ + \rho_- v_-$$.
3b. The positiv charges are not moving. Thus $$v_+=0$$ so that $$j_+=0$$. Thus, we obtain $$j_{tot} = j_+ + j_- = \rho_+ v_+ + \rho_- v_- = \rho_- v_-$$ Note, that we do not (!) imply $$\rho_+=0$$, as this would be wrong: We have $$\rho_+ = \rho_- \ne 0$$.