# A question on electric current

In my physics book the definition of current is

If a charge $$∆Q$$ crosses an area in time $$∆t$$, we define the average electric current through the area during this time as
$$I=\frac{∆Q}{∆t}$$. The current at time $$t$$ is $$I=\lim_{∆t\to0} \frac{∆Q}{∆t}=\frac{dQ}{dt}$$

And the definition of current density is

If $$∆I$$ is the current through the area $$∆S$$, the (same as the top) ... $$J=\frac{dI}{dS}$$

My question is what is the difference between the 1st definition area and the 2nd one?

• Use $\Delta$ instead of ∆
– Rew
Commented Jan 25, 2021 at 9:20
• And in fact, both the areas being talked about are the same! It's just a matter of how you want to calculate it; whether you want the current through only half the wire's cross-section or 3 quarters of it.
– Rew
Commented Jan 25, 2021 at 9:22
• It is the same area. Its symbol is only needed for the density equation, so doesn't appear in the one it is not needed for. Commented Jan 25, 2021 at 9:40
• Still I am feeling somthing. The first area can be anything, right? In the first one the area doesn't matter. But if dI=J.dS then area does matter and it could not be anything Commented Jan 25, 2021 at 9:49

The total current $$I$$ through a surface is a global quantity (with respect to the surface). The current density $$\bf J$$ is a vector and it is a function of the point, in general. The integral over the surface of $$\bf J \cdot n$$ is the total current $$I$$, $$\bf n$$ being a unit vector orthogonal to each surface element. If the surface is so small that variations of $$\bf J$$ over the surface can be neglected, and if the surface is orthogonal to the direction of $$\bf J$$ there is a relation between the small surface area $$\Delta S$$, $$|{\bf J}|$$, and the total current through the same surface: $$I=|{\bf J}| \Delta S.$$ If the unit vector makes an angle $$\theta$$ with $$\bf J$$ the previous formula becomes: $$I=|{\bf J}| \Delta S\cos \theta.\tag{1}$$ The right-hand side can be interpreted as the product of the modulus of the density current by the cross-sectional area of the surface element.
Summarizing: if $$I$$ is the total current through a small surface element $$\Delta S$$ and we want to connect it to an underlying current density $${\bf J}$$ the surface element must be the same, in general. However, formula $$(1)$$ tells us that many surface areas, oriented to give the same cross-sectional area, will give the same total current.