# Does voltage depend on the surface area of a conductor?

I am beginning to study the properties of electricity and have a question. I was told that electrical voltage is basically like pressure in a pipe. However, pressure in a pipe is a measurement of force per unit (pound per square inch) of surface area, which means if I take the same force and spread it out to double the surface area I should get a pressure of half of what it was for the same surface area.

Now from what I understand, voltage isn’t measured in volts per square inch or anything like that, so can someone explain why electrical pressure and water pressure don't share the same type of pressure measurement and where the confusion comes into play? If I connect a conductor with 10x the surface area to both sides of a battery terminal will I get a voltage drop of 1/10th of the voltage since the electric pressure (voltage) is being spread out on a surface 10x the size?

## 3 Answers

The analogy is good, but you appear to misunderstand it. Water pressure is not a function of the surface area of the piping, any more so than voltage is a function of the surface area of the wiring.

The water pressure is a function of the strength of the pump which is charging the pipework (assuming the quantity of water available to charge the pipework in the first place, exceeds the capacity of the pipework). And in a situation where the water is flowing freely out after the pump, it is also a function of the resistance of the pipework that comes after the pump.

A static pressure of 1bar in a small diameter pipe is the same as a static pressure of 1bar in a large diameter pipe. Drill a small hole in either pipe, as if applying a multimeter to a wire, and you will measure the same 1bar of pressure in both.

The difference is when the water is flowing. If the water is at a pressure of 1bar leaving the pump at the pump's outlet, then there will be a pressure drop along the pipe according to the cross-section (in the ideal case of a circular pipe, the diameter) of the pipe and the distance from the pump outlet. The pressure at the open end of the pipe will not be 1bar, but something less depending on the diameter and the length from the pump.

The drop will be larger for smaller pipes than larger ones.

However, if the open end of the pipe is then closed, the pressure will charge up to the same pressure as at the outlet of the pump, and the final static pressure will be the same regardless of the diameter of the pipe.

This is why "voltage drop" has to be measured with the relevant load switched on, just as pressure drop has to be measured with the outlet open.

As to units, the electrical analogue unit for water volume - often measured in litres - is the coulomb. The electrical analogue unit for water flow - not given a first class unit, but often measured in litres/second - is amps.

Pounds per square inch - psi - is slightly confusing in this context. It does not mean the pressure is a given number of pounds distributed over the available square-inchage of the pipework (so that increasing the square-inchage of the pipework leads to a reduction in the number of pounds applied to each square inch). The psi is the amount of pressure being applied to every square inch of the pipework, howsoever many square inches there may in fact be.

Increasing the internal surface area of a solid container does not reduce the pressure of the contents. But increasing the volume of a container requires more litres to be added to charge the container to a given pressure in the first place.

Similarly, the surface area of a wire does not change the voltage - the voltage is the pressure that is apparent across all surface of the wire, and increasing the surface area of the wire will simply mean the exact same voltage level being apparent across a larger surface than before.

What a wire of larger volume does do, is require more coulombs of electricity to charge the wire to that voltage in the first place. The eletrical analogue unit of volumetric capacity - also often measured in litres, confusingly (mainly because liquids are conventionally treated as incompressible) - is the farad.

Edit: I thought I'd add further to this as it provoked my own interest about the water analogy. I hope I haven't spoiled a good start.

I was wondering how to analogise watts and joules. I've already said that coulombs are the analog unit of water quantity in litres. We would usually measure water transfer or consumption in terms of such litres. But why do we very rarely talk of electrical quantities in terms of coulombs?

We more often talk of everyday electricity consumption in terms of watt-hours - which is an equivalent to joules (1 watt-hour = 3,600 joules). What then is the water analog unit for joules?

The answer is that it is actually joules for both. Joules (and by implication watt-hours) is a unit of energy or work quantity.

"Work" here is not a description of a physical quantity. It is a description of a purely abstract mechanical quantity, based on the ability of the system in question to bring about change in the physical properties or spatial arrangements of physical things. When a quantity of work is transferred in a system, we are describing what quantity of change has occurred in that system.

The word "change" here is intentionally elusive. Specific kinds of physical change can be measured in different units - a litre of water is raised off the ground by 1 meter, or a litre of water is heated by 100 degrees. But each of these specific changes can be reduced (by applying the laws of physics) to a general description of the quantity of work transferred.

Note also that I talk of "work transferred", not "work done". Because this quantity of work is neither created nor destroyed, nor produced or consumed, only transferred. And it can be transferred on (so as to cause consequential change), or transferred back (so as to undo the change), and the laws of physics describe the terms on which this can be done. And there are always two sides to the equation - if work is transferred to some change, there must be another change elsewhere of opposite work quantity.

Nowadays we rarely use transfers of water through pipework to do work in this mechanical sense. We pipe water into the home as a solvent for washing, as a carrier for sewage, or for human consumption, and it is the water itself (i.e. it's intrinsic physical or chemical properties) which is seen as the main feature for this purpose, not the work which the piping and the water transfer system is capable of performing.

Transferring water in and out of the home does perform work, but it is not of everyday interest to measure exactly how much work is performed in doing so.

The contexts in which water, or steam, are used specifically to do mechanical work, and where that work would be subject to measurement, are in the realm of mechanical engineering. A main application is in turbines (of which a simple waterwheel is but a simple example). Now we see an application in which the quantity of water in litres is no longer intrinsically relevant, but instead it's ability to do work by driving the turbine is the quantity of interest. An unpressured quantity of water is not capable of doing any work in driving the turbine. But a pressured quantity is. And the amount of work is a function of the pressure of the transfer, not just the volume.

And that brings us back around to electricity. We do not move eletrical charges around wires for the intrinsic properties conveyed by charge, or so that we can hold the charge itself in our hands. We do it so that the movement of electrical charge performs work - usually in the end, work of a non-electrical kind, such as heating things or driving mechanical appliances. And that is why we measure electrical consumption in terms of joules, or watt-hours, because the ability to transfer work (rather than to transfer electrical charge) is the main purpose of electrical systems.

And finally, what of the watt unit? That is a measure of power, or the rate of transfer. But crucially, it is not a measure of the rate of transfer of litres of water, or of coulombs of electrical charge. It is a measure of the rate of transfer of work (measured in joules), the abstract quantity which I've described above.

It is not, as might first be thought, like the transfer of litres of water. A transfer of a very small amount of water under very great pressure, is capable of doing the same amount of work, as the transfer of a very large amount of water under more modest pressure. And watts measures the rate of work transferred electrically, regardless of the voltage or amperage used to do it (and thus regardless of the amount of coulombs actually moved, which in AC circuits is a net of zero anyway - it's like the same water volume being pumped and sucked back and forth in rapid alternation).

@Steve's answer is really very very good, and I can't improve upon it. But I want to add my thoughts anyway.

I find this analogy very useful, not just for helping understand why the currents in various legs of a circuit are what they are, but also for building intuition about what voltage actually is and where it comes from.

Water under high pressure really wants to escape--if there's any path to get to somewhere where the pressure is lower, it will take it. It's the same with electric charge--if it's somewhere where the voltage is high, it really wants to escape. That's why it's useful to think about water pipes to figure out where current will flow.

But $$why$$ does the water want to escape? What is physically different about it than if it wasn't at high pressure? The reason is that it's squeezed. Water molecules like to be so close together and no closer, and when they are a bit closer than they like, they push apart. Molecules of water under high pressure are a bit closer together than usual, and when you see pressurized water come spraying out of an opening, the force that's causing that is the molecules trying to spread back out to their desired separation. It's just the same with charge and voltage: the charges in wire at high voltage are bunched ever so slightly closer together than usual (or further apart--there's both positive and negative charge but let's not worry about that). Just like the water molecules, they want to separate back out, and that's what it is about them that makes them different from charges in another wire that's not at high voltage.

Every analogy has its limits, but if you find water pressure and flow easy to think about, this one can take you a long way, in my opinion.

As for your question about why the units of pressure are force per area while the units of voltage are just volts: it is true that pressure can be expressed as force per unit area, but maybe more aptly for this analogy, it can also be expressed as energy per unit volume: joules per meter cubed is the same thing as newtons per meter squared. So you can see that the pressure of a given amount of water tells you how much energy you can get from it by letting it flow to where it wants to go. In exactly the same way, voltage is the amount of energy you can get from a given amount of charge by letting it go where it wants. The amount of charge is not measured by volume (otherwise voltage would actually be pressure) but both voltage and pressure can be thought of as "energy per amount of stuff" (where stuff is either water or charge).

• You say not to worry about positive and negative charge, but it actually very easy to extend the water analogy to a pipework circuit, where a pump induces not only pressure at its outlet, but also suction at its inlet at the opposite end of the circuit. Even in the absence of an explicit pipework circuit, pumps do in fact function in this way - if you pump water in one direction from an open tank, the air pressure above the water in the source tank drops, and this eventually sucks air all the way back from wherever the water is being deposited. Feb 9 '21 at 5:35
• (or further apart--there's both positive and negative charge but let's not worry about that) I just inserted so many mental caveats¹²³ Feb 9 '21 at 21:06

That's a really bad analogy in my opinion. It's much better to understand what electric potential difference, colloquially called voltage, actually is.

From electrostatics, you might know that the electric potential at a point in an electric field is the electric potential energy per unit positive charge at that point. If there is a difference in electric potentials and thus the E.P.E of a unit positive charge between 2 points, the electric force must have done some work on the charge. A little mathematics shows you that the electric field strength (for per unit positive charge) = -dV/dx where V is the electric potential. Electric field strength is the negative derivative of electric potential and ALWAYS points in the direction of decreasing potential.

This means that wherever there is a potential difference between 2 points in a circuit, there is an electric field between them pointing in the direction of decreasing potential and positive charges from the higher potential to the lower potential. It is here where the greatly misleading analogy of "electric" pressure comes from. The only thing even remotely similar is that both a potential difference and pressure result in a force that pushes charges/ water molecules.

This analogy is just to give beginners some intuition that voltage causes charges to move. If you're comfortable with electrostatics, however, its much better to know that the potential difference creates an electric field and it is this electric field that exerts a force on charges and causes them to move.

• Steve's answer excellently elaborates why the pipe analogy is referred to in the first place, and realistically discusses its limitations. My answer hinted at how to actually calculate current and voltage for the most basic geometries, as the OP desired. I personally think your explanation is a confusing to beginner, and inadequate to guide the intuition of the OP. Feb 9 '21 at 3:38
• Your answer is ill-informed and doesn't even answer the question. All you have done is stated a bunch have well known formulas without explaining how voltage varies and what is kept constant. Op clearly asks if connecting a wider conductor between the terminals will decrease the voltage. Voltage here is not a function of resistivity and your answer misleads gravely Feb 9 '21 at 3:55
• The only reason I did not explain any more is that I have done so in a detailed fashion multiple times on recent questions, and indicated so in my answer. And, to be clear (to you), of course voltage drop across circuit elements depends on their resistance, once you start putting circuit elements together in series and applying a voltage across them all. Feb 9 '21 at 4:15
• Voltage depends on resistance in the case of multiple components in series, whereas op clearly mentioned connecting a conductor to both sides of a battery, so it appears you didn't even read the question. Moreover, your formulas without any context imply that the voltage drop also depends on the current, which it obviously doesn't Feb 9 '21 at 4:44