# Current Definition

1. Feb 11, 2016

When the current is defined as being the conventional current then:
i = dq/dt, i = integral of J*ds

When the current is defined to be the electron flow:
i = -dq/dt, i = - (integral of J*ds)

Is this right?

2. Feb 11, 2016

### nasu

If you define the current in terms of negative charge, won't you do the same for current density? Why the - sign in front of the integral?
You define J with positive charge and i with negative?

3. Feb 13, 2016

### vanhees71

You don't need to think in this complicated way about currents. Just use vectors! The current density
$$\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x}),$$
where $\rho$ is the density of electric charges and $\vec{v}$ is the flow-velocity field of the charged matter. Now everything is encoded in this equation. Particularly the sign of the charge density determines automatically whether the current density is pointing in or opposite to the direction of the charge flow velocity.

The sign of the current then is uniquely defined by the choice of orientation of the area this current is referred to: Electric current is the amount of charge per unit time flowing through a given area. It's orientation is defined by the choice of the surface-area element vectors, perpendicular to the surface, $\mathrm{d}^2 \vec{f}$. Then the current is uniquely defined by
$$i(t)=\int_{F} \mathrm{d}^2 \vec{f} \cdot \vec{j}.$$
Keeping these fundamental definitions of the quantities in mind there's no more confusion about signs and you don't need vague descriptions like "conventional current" anymore.
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