# Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section

Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section?
Before I explain my query, I would like to clarify that I am a ninth-grader who got this question while studying the formula $R\propto \frac{1}{A}$ where $A$ is the area of cross-section.
I have often asked this question to my teachers and they always give me the classic "corridor and field example". They told me that if 5 people walk in a corridor, they will find it harder to get across than if they were to be walking through a field- the same goes for electrons passing through a conductor. My counter-argument would be that if the width of the conductor increases, so will the number of positive ions (my textbook says that positive ions in conductors hinder the flow of current) and hence, more the resistance.
I would really appreciate it if the answer could be explained to me in simple terms as I'm not well versed with the more complex formulae involved in this concept. If not, do let me know of the concepts I should read about (preferably the specific books) to understand the solution better.
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Eliezer Olson
Sorry for my poor english. My native language is french.
We can also ask the question about the conductance: why is it proportional to the surface.
The reason is that we are working with a model in which the current density is distributed uniformly over the cross section of the conductor. It is as if we had a lot of identical conductors in parallel with each other. So, for the same voltage, if we multiply the area by N, we also multiply the current by N. This would not be the case if the current distribution was not uniform. For example, if we take into account the skin effect.
Another interesting example is that of the Hagen-Poiseuille's law in hydraulics: the flow rate is proportional to the pressure difference but the hydraulic conductance is proportional to the square of the area: it is not proportional to the area ! This is because the velocity profile is parabolic (and not uniform). Doubling the pipe area is not the same as taking two pipes.
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Brooklynn Hubbard
The following model is not entirely true, but it's a much truer model than people walking down a corridor. I hope it helps.
Start by considering a resistor that just one atom wide, but many atoms long, in a long string.
Each atom has an electron cloud. In a semiconductor, the outermost electrons in the electron cloud are weakly bound to their atoms. It doesn't take a very strong push to separate the electron from its atom, but it's not free, either. And pushing an electron free from its atom doesn't make it free to keep moving that electron.
As soon as the electron is pushed out of the first atom it will see another atom, sitting there with all of its electron orbitals filled. It needs to push one of that atom's electrons out of its orbital and take its place. Then as soon as that electron is pushed over, it encounters an atom, and so on.
The end result is that if we want to push an electron out at one end, we need to push an electron in at the other end, with an amount of force equal to the total amount of force required to push all the electrons one atom over. We pop one electron in at one end of the resistor, it pops an adjacent electron one atom over, that one pops the next electron one atom over, and so on, until we get to the end of the resistor and an electron pops out.
In a circuit, the circuit voltage sets the amount of force that the circuit can provide, per electron.
So, let's say that the circuit voltage can push ten times harder than is needed to push one electron through one line of atoms in one second.
Then the circuit can shove ten electrons a second into one end of the resistor, and ten electrons a second will fall out of the other end of the resistor. Current is flowing... but not very much.
What if we add a second line of atoms in parallel with the first? The circuit pushes per electron, so it can push another ten electrons per second through the second line of atoms. Two lines of atoms means two times the current.
So on for each additional line of atoms: each line of atoms gets us another ten electrons per second getting shoved in on one end and another ten electrons per second falling out on the other end of the resistor.
The cross-sectional area of the resistor is proportionate to the number of lines of atoms we can shove electrons in one end of so that electrons fall out the other end.
So we can say: $A\propto I$, the cross-sectional area is proportionate to the current.
And since $R=V/I$
$R\propto 1/A$
We can go farther with this model and reach the standard formula for resistance.
What happens if we add more atoms to each line, in series, so that the line of atoms is longer? Each atom opposes our push by the same small amount, and is the same size. So, the resistance must be proportionate to length.
$R\propto L$ and $R\propto 1/A$
so
$R\propto L/A$
Now we just need a proportionality constant related to how hard each electron is to push one atom over. For Resistance, the proportionality constant is represented by $\rho$ (rho), the material's Resistivity.
$R=\rho L/A$