This is a confused part ever since I started learning electricity. What is the difference between electric potential, electrostatic potential, potential difference (PD), voltage and electromotive force (EMF)? All of them have the same SI unit of Volt, right? I would appreciate an answer.
[Physics] the difference between electric potential, electrostatic potential, potential difference (PD), voltage and electromotive force (EMF)
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Before thinking about circuits, let's think about two conducting spheres of charge that I connect by a wire. Before I connect the wire, sphere 1 is at voltage $V_1$ and sphere 2 at voltage $V_2$, let's say $V_1>V_2$. I find it useful, in terms of thinking about what's going on, to notice that if the spheres are the same size then saying the spheres have different potentials is equivalent to saying that the 2 spheres have different charges residing on their surfaces (you can justify this by noting that the capacitance of a sphere is determined by its radius).
Now let's connect the spheres. What will happen? Well, a current will flow in the wire. This will take positive charge off of sphere 1 and deposit it on sphere 2 [strictly speaking if you want electrons to be charge carriers, then negative charge is flowing from 2 to 1; but in terms of thinking about what's going on it's easier to imagine, and mathematically equivalent to say, that positive charges are going from 1 to 2]. This in turn changes the voltages on the two spheres; $V_1$ decreases and $V_2$ increases. The process stops when $V_1=V_2$. Again, if the spheres are the same size this condition is equivalent to the charges on both spheres being equal.
OK, now imagine a battery hooked to a resistor and a switch, the simplest circuit imaginable. Before we close the switch, terminal 1 is at $V_1$ and terminal 2 is at $V_2$. At this point, it makes perfect sense to think of each terminal of the battery as being a sphere of charge. Then we close the switch, this is like connecting our spheres with a wire. Based on our silly model of a battery, you would the voltage between the two terminals of the battery (ie, $V_1-V_2$) to decrease until eventually it reached equilibrium with $V_1=V_2$. Clearly, a battery does not behave like two spheres of charge after the circuit is closed.
The whole point of a battery is that it maintains the potential difference between its two terminals. After we close the switch, a little bit of positive charge flows from terminal 1 to terminal 2 by going through the circuit. Naively this means that terminal 1 has less positive charge and terminal 2 has more positive charge, so terminal 1's voltage decreases while terminal 2's voltage increases. Inside the battery, some process takes place to to take the excess positive charge on terminal 2 and put it back onto terminal 1. Whatever this process is, it cannot be electrostatic, because positive charges following the electric field can only ever move from terminal 1 to terminal 2 [positive charges move from high voltage to low voltage, if the only force is electrostatic].
The details of what the battery does to maintain the potential difference varies depending of the kind of battery. A conceptually simply example of a battery is a Van de Graaff generator. In a Van de Graaff generator, you have a conveyer belt that literally carries the excess positive charge on terminal 2 and deposits it back on terminal 1, undoing the naive 'equilization process.'
Most useful batteries rely on some chemical process to maintain the potential difference. For example, one can use oxidation reactions to do this. The details involve some chemistry (there's a wikipedia summary at http://en.wikipedia.org/wiki/Electrochemical_cell), but essentially you put each terminal in a bath of ions, and the chemical energy of the reactions at each terminal [balancing oxidation and reduction] forces ionized atoms to carry electrons from terminal 2 to terminal 1.
The distinction between emf and potential difference is often glossed over and often misunderstood so this is an appropriate and interesting question.
Since both emf and potential difference are measured in volts, it is quite easy to use the terms interchangeably and, in many cases, there's no harm done but that fact is that emf and potential difference are distinctly different concepts.
Essentially, it is the emf that moves charge around a closed path - a circuit - while potential difference is due to the charge distribution along the circuit.
Consider the first circuit; there is a battery (or cell actually), the source of emf. If there is no circuit connected to the battery, the potential difference between the battery terminals is equal in magnitude to the emf.
From the Wikipedia article "Electromotive force":
In the case of an open circuit, the electric charge that has been separated by the mechanism generating the emf creates an electric field opposing the separation mechanism. For example, the chemical reaction in a voltaic cell stops when the opposing electric field at each electrode is strong enough to arrest the reactions.
The electric charge that has been separated creates an electric potential difference that can be measured with a voltmeter between the terminals of the device. The magnitude of the emf for the battery (or other source) is the value of this 'open circuit' voltage.
If an external circuit (path for charge to flow) is connected to the emf source, electrons from the more negative terminal will flow through the circuit to the more positive terminal.
There is no better demonstration of the distinction between emf and potential difference than to place a short circuit across the terminals of the battery (this must be done carefully and quickly) to determine the short-circuit current.
Assuming an effectively ideal short circuit, the potential difference between the battery terminals will be effectively 0V yet there will be a large current through the short-circuit.
It is the emf that 'drives' this current round the closed path.
Now, once again consider the first circuit. Clearly, the emf is driving the electrons clockwise but it is also clear that, for there to be a current through the resistor, there must be a potential difference (Ohm's law). Thus, the electron density is slightly greater on the conductor connected to the more negative terminal of the resistor than on the conductor connected to the more positive terminal.
There's much more to emf than what I've outlined here and the Wikipedia article (and links and reference therein) I linked above gives much more detail.
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EDIT: Put simply, potential difference is the work done by electrostatic force on a unit charge, while EMF is the work done by anything other than electrostatic force on a unit charge.
I don't like the term "voltage". It seems to mean anything measured in volts. I'd rather say electric potential and electromotive force.
And the two are fundamentally different.
Electrostatic field is conservative, that is, over any loop $l$ we have $\oint_l \vec{E}\cdot\mathrm{d}\vec{l}=0$. In other words, the line integral of electrostatic field does not depend on the path, but only on end points. So we can define point by point a scalar value electrostatic potential $\varphi$, such that $$\varphi_A-\varphi_B=\int_A^B \vec{E}\cdot\mathrm{d}\vec{l},$$
or
$$q \left( \varphi_A-\varphi_B \right)=\int_A^B q\vec{E}\cdot\mathrm{d}\vec{l},$$
so $q\Delta\varphi$ equals the work done by electrostatic force.
In pratical application, electrons (and other carriers) flow in circuits. Since electrostatic field is conservative, it alone cannot move electrons in circles; it can only move them from lower potential to higher potential. You need another kind of force to move them from higher potential to lower ones in order to complete a cycle. This other force could be chemical, magnetic or even electric (vortex electric field, different from electrostatic field), and their equivalent contribution is called electromotive force. $$\mathrm{E.M.F.}=\int_\text{Circuit} \frac{\vec{F}}{q}\cdot\mathrm{d}\vec{l}$$