# Question about the relationship between current density and charge density

In wikipedia the two concepts are related as(in Magnetostatics ):

$$\vec j(\vec r)=\rho(\vec r)v$$ where $$v$$ is the velocity with which the particles move.

I consider a volume in which initially there is no charge and initial time is $$t_0=0$$ and $$Q(t_0)=0$$ I tried to derive the above formula the following way:

$$\frac{dQ}{dt}=I$$

$$\int dQ=\int I dt$$

$$Q=It$$

$$\rho(\vec r)V= I t$$

$$\rho(\vec r) \cdot A \cdot l= j(\vec r )\cdot A \cdot t$$ ,where $$A$$ is the surface

$$\rho(\vec r) \cdot \frac {l}{t}=j(\vec r )$$

$$\rho \cdot v=j(\vec r )$$, where $$v$$ as said above, is the velocity of the charged particles.

If this derivation is correct, my question is:

Is the inital formula, only when the current density is not time dependent? Is there a more general formula for when $$\vec j= \vec j(\vec r,t)$$ ?

The most general equation for charge continuity is

$$\nabla\cdot \vec{j} = -\frac{\partial\rho}{\partial t}$$

Which is true for all points $$\vec{r}$$ and all times $$t$$.

Usually the general definition of the current density is

$$\vec j: \mathbb{R}^3 \times \mathbb{R} \to \mathbb{R}^3 \\ (\vec r,t) \mapsto \vec j(\vec r, t) := \rho(\vec r, t) \vec v(\vec r, t)$$

This can be used to show that

$$I(A) = \int_A \vec j \cdot \, \mathrm{d} \vec S$$

In your derivation you are using

$$\int I dt = I t$$

which is only true for constant currents. Furthermore you use $$V = Al$$, this implies that $$A$$ is the ground plane, but on the other side of the equation it is the surface.

• $A$ is the same. If you consider a wire of cross section $A$, and you consider a volume in this wire, that's the result of $A \cdot l$ the distance covered by the particle in $A$ after some time t. And in the right side I am using the same surface idea, to calculate the total current that passes through this cross section during the time t Commented Feb 10, 2022 at 1:18
• @imbAF If you consider a volume of a wire, then the change in charge would be 0, because the current goes in one side and out on the other. What you mean is the charge that is transported through the cross-section of the conductor, in which case your calculation is correct. The point is that the charge that is transported through the cross-sectional area is exactly the same as the charge in a volume with the same area and a length l= v*t. Commented Feb 10, 2022 at 8:09