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How can people say that voltage is like a pressure. They also say that voltage can be seen as on how badly the electrons move. From my knowledge, voltage is the work done per unit charge and so if the work done is high then the voltage / potential difference is high. Also, what do we mean by per unit charge (is it 1 coulomb charge, or what should it be)? Tnx

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Voltage is like pressure in a hydraulic or pneumatic system. If you have a small diameter pipe (or a high resistance wire), you will need more pressure (voltage) to get the same flow (current). Also, with more pressure (voltage) each unit of steam (charge) does more work.

By unit charge we mean some fixed charge, usually it is 1 coulomb, but for relative calculations you can take any fixed charge (say, 1 electron or 4200 electrons).

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    \$\begingroup\$ Also, power == pressure * volume-rate-of-flow == voltage * current \$\endgroup\$ – MikeJ-UK Apr 2 '12 at 13:41
  • \$\begingroup\$ regarding electrons-by-the-dozen: just take care to divide by the charge as measured in coulombs; 1.602e-19 coulombs each. \$\endgroup\$ – JustJeff Apr 3 '12 at 0:43
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A note on the possible misunderstanding: you said that

voltage is the work done per unit charge and so if the work done is high then the voltage / potential difference is high.

This is false: to build the voltage, the charges don't have to do any work. Their simple existence is enough to create an electric field in the surrounding space; this field will have a maximum (in magnitude, so can be a minimum if the charge is negative) and will go to 0 at infinite.

The electric potential is related to the potential energy that a charge has:

$$ U_{E} = qV $$

so a charge q at a potential (absolute) V will have a potential energy Ue; it'll tend to give energy (in the form of charge, in this case) to a point with lower potential, and take energy from a point with higher potential.

The difference of potential will be given by the integral of the electric field between the two points.

This was only to say that you can just spread your arms and the tip of your fingers will have a voltage between them.

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  • \$\begingroup\$ I don't think that's right. If one electrode has a higher potential than the other (meaning it has an electron shortage), the only way electrons will move from the former to the latter is if work is done to make them. Conversely, electrons will want to move from the latter to the former, even if they have to do work to get there. If one moves a coulomb of electrons between two plates that are close together, and then separates them, one can build voltage without moving charges, but such separation will require mechanical work against the electric forces attracting the plates. \$\endgroup\$ – supercat Apr 2 '12 at 16:14
  • \$\begingroup\$ @supercat I don't understand what's wrong: potential doesn't mean necessarily work (in the physical sense). If you have a potential, the only way to prevent charge from moving is to put the vacuum in between; otherwise, the charge will tend to spread uniformly (such as heat). Then, if you want to create a usable potential, you have to force this mechanism such as opposite charges are accumulated in specific points. \$\endgroup\$ – clabacchio Apr 2 '12 at 17:49
  • \$\begingroup\$ When a potential exists, electrons want to flow from the more-negative point to the more-positive one. If the impediments to flow aren't do great, the electrons will do work, even if only by heating up the wires through which they are traveling; such work will be proportional to the number of electrons and the potential difference of the points between which they travel. \$\endgroup\$ – supercat Apr 2 '12 at 17:58
  • \$\begingroup\$ @supercat fine; but this doesn't explain why you need work to have a potential \$\endgroup\$ – clabacchio Apr 2 '12 at 18:08
  • \$\begingroup\$ One can theoretically create a potential difference without doing work, if one isn't moving any charges; in practice, one can sometimes create a massive potential difference by moving a sufficiently small amount of charge that very little work is required. On the other hand, if that same tiny amount of charge is allowed to flow back where it was, the massive increase in potential will be undone without doing much work. \$\endgroup\$ – supercat Apr 2 '12 at 18:40
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Voltage is conceptually very much like pressure: the plumbing/pipes analogy for circuitry, the Hydraulic Analogy.

Pressure is force per unit area. And for a charged surface, the potential-difference over a fixed length also gives a force per unit area. Increasing the voltage gives greater "push." This gives us Ohm's law: if "friction" is constant, then higher voltage gives faster charge-flow, so higher currents, and V/R=I

Voltage along a resistive wire is a lot like pressure along a narrow pipe full of viscous fluid. If we double the pressure-difference at the ends of the pipes, the fluid flow rate will double. And doubling the voltage across a resistor will double the current.

A voltage between a pair of parallel plates creates a certain force, larger for plates with more surface area. This is much like the pressure-difference between two sides of a piston inside a cylinder.

Voltage is also like pressure in that, at a single point, pressure is meaningless, and so is voltage. In other words, a single point can have many pressures at the same time, depending on which distant point we pick as a reference pressure. And, the voltage on a wire is meaningless, since a wire can have many voltages at the same time, depending on which second wire we choose as our zero-volt reference. Pressure and voltage must be measured between two locations.

Of course the pressure-analogy of voltage needs a charged surface, and this shows how voltage differs from pressure: e-fields and their voltages are 3D effects, while pressures are instead surface phenomena. For charge distributed inside a 3D object, voltage can apply a force to the inside of the object and drag the entire charge along. Pressure only applies a force to a surface, and is proportional to the surface area, and not to the volume or mass.


Also note:

Yes, the unit called "Volt" can be defined as the work done per unit charge. But be very careful with that.

Why? Because Electrical Potentials themselves, "voltages," are quite different: they're numbers associated with e-fields. E-fields are not just "work done per unit charge." We can have e-fields even when no "unit charge" or potential energy is present. Defining voltage as "work per unit charge" is like defining "gravity" as just being the work done per kilogram being lifted. Yet if we take away the kilograms of boulder, the gravity and Gravitational Potentials are still there in space! Gravity is not work done per kilogram, and voltage is not work done per unit charge. "Potential" is not "Potential Energy." Instead, "Potential" is "distance in a field of particular field-strength." Only the field is needed for describing Potentials; no charges or PE.

On the other hand, our unit called "Volt" is directly associated with the units for charge and energy. Very confusing! Heh, "Volt" is defined by potential energy, but "voltage" is not! So, it's probably more sensible to work backwards: if we know what are the Charge and Volts, then "Potential Energy" can be defined as the energy required to transport a charge across a difference in voltage. If we need to know the potential energy, first we measure the voltage-difference and the coulombs.

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