Conventional current definition and a variation on that definition

[ChiralSuperfields]

Homework Statement

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Relevant Equations

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One normally sees that the definition for conventional current as defined as the amount of positive charge that passes a point over unit time. However, why could we not define conventional current as the amount of positive unit charge that passes a point over unit time.

I added in unit there since the definition of conventional current holds true for any unit we choose measure charge. However, it appears that there is only a metric unit for charge.

Any thinking about what I have said is greatly appreciated.

Many thanks!

 

[haruspex]

I fail to see that inserting the word unit achieves anything desirable.
In " the amount of positive charge that passes a point over unit time" (or, as I would phrase it, " the amount of positive charge that passes a point per unit of time"), the word unit is not being used in quite the same way as in "seconds are units of time". Seconds are standard units of time, as are years, etc., whereas defining a rate as "an amount per unit of time" does not imply standard units, let alone any particular standard unit.

Would it bother you less if worded as " the amount of positive charge that passes a point per length of time"?

(Just struck me that it ought also specify "in a given direction", otherwise specifying positive is meaningless.)

 

[pbuk]

However, why could we not define conventional current as the amount of positive unit charge that passes a point over unit time.

Because that would not make sense: the amount of charge in a unit of charge is one unit.

The two statements below are equivalent, but you can't mix bits of them together:

• "[The amount of] Conventional current is the amount of positive charge that passes a point over unit time"
• "One unit of conventional current is one unit of positive charge passing a point in one unit of time"

Neither is a particularly good definition; I would rather adapt Wikipedia's definition of electric current to give "Conventional current is the net rate of flow of positive electric charge through a surface".

 

[vanhees71]

The problem with the notion of "current" is that it is one of the concepts which is (a) less simple than thought on first sight and (b) often not well explained particularly in introductory textbooks. The reason is that for some (historical?) reason textbook writers think the integral form of Maxwell's equations were the most simple starting point although this is not the case.

It is important to first making clear that a current is a scalar (!) quantity giving the amount of charge per unit time flowing through a given surface. The sign depends on both the direction of the corresponding current density, which is a vector field (!) and the arbitrarily chosen direction of the surface normal vectors across which this current density is integrated to define the current.

This indicates that, as usual, the local notion of the phenomenon under consideration ("moving electric charge") is simpler than the global (integral) one. Starting with the charge density (a scalar field), $\rho(t,\vec{r})$: if you take a small volume $\mathrm{d}^3 r$ around the position $\vec{r}$, then $\mathrm{d} Q=\mathrm{d}^3 x \rho(t,\vec{x})$ is the charge contained in this volume at time $t$. Then if this charge is a fluid described by the velocity field $\vec{v}(t,\vec{x})$, then the current density simply is $\vec{j}(t,\vec{x})=\rho(t,\vec{x}) \vec{v}(t,\vec{x})$. Here $\vec{v}(t,\vec{x})$ is the velocity at time $t$ of a fluid element being at position $\vec{x}$.

Now the current for a given surface $A$ is given by
$$I(t)=\int_{\mathbb{A}} \mathrm{d}^2 \vec{f} \cdot \vec{j}(t,\vec{x}).$$
It's positive (negative) if the flow of the charge is more in (opposite to) the direction of the surface-normal vectors $\mathrm{d}^2 \vec{f}$. That's all. There's no need for a "conventional current" and a "real current". The signs are all be taken of by these mathematical definitions of the current density vector field, which is of course in the opposite direction than the velocity field, if the charges are negative (i.e., if $\rho<0$). In addition it's important to keep in mind that the sign of the current also depends on the choice of the direction of the surface-normal element vectors along the surface, the current is referred to.

 

[ChiralSuperfields]

Thank you for your replies @haruspex , @pbuk and @vanhees71!

Many thanks!