In my textbook, Current is defined as the rate at which charge flows through a given surface (cross sectional area). These are my questions:
How can there be current through a given cross sectional area? If the volume is 0, then how exactly is a current "flowing" through that area?
Apart from this, in $ I = Q/t $ and in $ I = dQ/dt $ what does time mean here?
Suppose you work for a branch of the government that worries about how many cars are in transit at any given time. So your boss gives you a stop-watch and a counter and he tells you to measure the flow of cars at particular points in several roads scattered around town. The idea is to map car flow through various points in order to direct further investigations of whether a particular track of road needs more stringent speed limits or the construction of more lanes.
In order to do that you take the "finish-line" approach. You're going to stay stationary next to a road with your stop-clock and your counter. So you start the watch and every time you see a car pass you by you hit the counter one up. Notice that you don't have extension along the road, the car need not take a finite amount of time to cross your detection zone, it need only pass you by. Which it could do regardless of how thin you are.
After a certain amount of time $\Delta t$ has elapsed, you stop measuring. The number of cars is your $\Delta Car$ and the measure $j=\frac{\Delta Car}{\Delta t}$ is the flow of cars through the finish line. Let's keep that definition in mind.
Then you go to your next spot and you realize that this new stop is a four-lane road with two lanes going in the same direction. This new road is a mess, it has way more traffic than the last one and you would expect for cars to go slower and for a correspondingly lower flow of cars through your finish line. So you measure $j$ and it.. is about the same as the one in the previous measurement, which leaves you baffled for soem time.
So what gives? The problem is that in the previous example only one car could ever cross the finish line at any given time, because there was a single lane in the direction of measurement, a single possible trajectory for a car to follow. In the second example, we have a situation like this:
Here your position is marked in red and the finish-line is the red line crossing the lanes. The green arrows mark the possible trajectories through which a car can cross the finish line and the yellow marks the intersection points between the trajectories and the finish line (which is a measure of the amount of possible trajectories).
So at any given moment you might count two cars going through the finish line at the same time, meaning that you can potentially get the same flow even if the cars are going at half the speed of the first example.
In order to solve the problem you want to know the flow by lane so you take your measurement of car flow $j$ and you divide it by the number of lanes $\ell$ such that you have a new quantity $J=\frac{\Delta Car}{\ell\Delta t}$ This quantity the "flow density" and the quantity $\ell$ is the number of yellow points in the above figure.
If you have a larger detection area, it does not matter if $\Delta Car$ is very high because it means all cars are spread across multiple lanes so the flow becomes smaller to account for the fact that the cars are moving slower.
We can make the analogy with the wire now. The flow of cars $j$ is the current $I$ if the cars are the charges $Q$. The two-dimensional possible trajectories (two-dimensional because in a road with multiple lane a car can change lanes many times before crossing) crossing a finish-line are now 3D paths crossing a "finish-area" like the one shown below.
In the car example there were discrete numbers of possible paths a car could take because there were a discrete number of lanes, but electrons are not constrained like that. They can trace trajectories that can cross any point in the area such that the number of possible trajectories is the number of points in the area, which is just the area itself. Which means that the current density $J$ is just the flow $j$ divided by the extension of the finish-area. Since our "charge flow" $j=\frac{\Delta Q}{\Delta t}$ is the definition of electric current $I$, we have that the current density is $J=\frac{I}{A}$. $J$ is the flow of electrons through the finish-area A.
Hopefully by now you can see that if $I=\frac{\Delta Q}{\Delta t}$ then $$ \lim_{\Delta t \to 0} \frac{\Delta Q}{\Delta t} = \frac{dQ}{dt} $$
Suppose that at a given time there are several charged particles (each with, say, 1 coulomb of charge) to the left of your chosen cross-section. Two seconds later, three of those particles have moved past the cross-section and are now to its right. Then $Q=3$ coulombs, $t=2$ seconds, and therefore the current is $I=Q/t=(3/2)$ coulombs per second.
First of all, volume is not zero! Volume = cross section area x length of conductor
Secondly; the length of the conductor is not part of the definition, so there is no need to refer to volume
