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The current density is defined as:

$$\textbf{J} (\textbf{r},t) := \rho(\textbf{r},t) \cdot \textbf{v}(\textbf{r},t)$$

where $\rho (\textbf{r},t)$ and $\textbf{v}(\textbf{r},t)$ is the charge density and the particle velocity at point $\textbf{r}$ and at time $t$. Based on that definition, how could one derive the following relation?

$$I = \iint_{\mathcal{A}} \textbf{J}(\textbf{r},t)\cdot d\textbf{S}$$

where $\mathcal{A}$ is a surface. I am asking this because authors tend to define the current density $\textbf{J}$ through the latter equation (having the current $I$ predefined). I wish to understand how the two are equivalent.

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    $\begingroup$ Isn't this the definition of current if $\mathbf{J}$ is defined like you say? $\endgroup$ Oct 4 '19 at 22:27
  • $\begingroup$ Nice one. But let's assume that $I$ is separately defined as: $$I = \frac{dQ_{S}}{dt}$$ where $Q_S$ is the amount of charge passing through an area $S$. $\endgroup$
    – Heath
    Oct 4 '19 at 23:17
  • $\begingroup$ Thanks for clarifying that point by the way. It is really important. What I am asking here is for a proof that demonstrates the equivalence of the two definitions. $\endgroup$
    – Heath
    Oct 4 '19 at 23:21
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The current density is:

$$\textbf{J} (\textbf{r},t) = \rho(\textbf{r},t) \textbf{v}(\textbf{r},t)$$

Let $\Delta Q(\textbf{r},t)$ be the charge that crosses the infinitesimal surface $\Delta S$ in time $\Delta t$. You can clearly see that the amount of charge that crosses is the amount of charge present in the volume $\Delta S \textbf{v}(\textbf{r},t)\Delta t \cdot \hat{\bf n}$, $\hat{\bf n}$ being the normal to $\Delta S$. I can then write $$ \frac{\Delta Q(\textbf{r},t)}{\Delta t} = \rho(\textbf{r},t) \textbf{v}(\textbf{r},t)\cdot \hat{ \bf n} \Delta S = \textbf{J}(\textbf{r},t)\cdot \hat{\bf n} \Delta S$$

Simply integrating over the surface we get

$$I(t) = \frac{dQ_S(t)}{dt}= \int_{\mathcal{A}} \frac{dQ(\textbf{r},t)}{dt} = \iint_{\mathcal{A}} \textbf{J}(\textbf{r},t)\cdot d\textbf{S}$$ .

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