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EMF of an EMF source (a battery for example) is defined as the work done by the non-conservative force(s) on charged particles as it passes through the terminals of the source divided by the charge of the particle. That is $\epsilon=\frac{dW}{dq}.$ This definition is very similar to that of potential difference. But the major difference, among others, is that the force, in this case, is of non-conservative nature while the potential function is defined only for conservative forces only.

The potential difference between two points is independent of the charge of the particle ie the same potential difference will be calculated between two points whether we calculate it using a $10C$ or $-1.9C$ charge and also independent of the path the charge take. But this is not the case for EMF. The EMF depends on the path and the charge on the particle as it is defined for a non-conservative force. Hence EMF of a source calculated using $10C$ particle along a path will be different if the particle took another path and it will be different still if we move $-1.9C$ charged particle along the same path. So what do we mean when we say that a battery has EMF $\epsilon$? How can we be sure that the work done by the battery-by the non-conservative forces due to battery-on a charged particle $Q$, as it moves through the source, will be $Q\epsilon$? Won't it be different if it took other paths?

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"The emf depends on the path and the charge on the particle."

The emf is indeed path-dependent, but not in general charge-dependent. The work, W, done on a charge, $q$, taken along a specific path is proportional to $q$ and so the emf, defined as $W/q$, is independent of $q$. Examples include (1) a wire of length $l$ moving at speed $v$ at right angles to itself and cutting magnetic flux, for which $\mathscr E =\frac {q(\mathbf v \times \mathbf B).\mathbf l} q =Blv$ and (2) a battery. For the battery the terminal voltage does depend somewhat on the curent, that is the charge passing per second through the battery, but we attribute this to the battery's internal resistance rather than to changes in its emf.

What about path dependency? In the case of the battery, a charge $q$ taking any path from terminal to terminal through the battery, that is through the interfaces between electrodes and electrolyte will result in the same value for $W/q$. [For example, for each pair of electrons that enter and leave a simple cell, one zinc atom will be ionised and two hydrogen ions will accept electrons from copper, so a particular net amount of energy will be released.] But if we take $q$ from terminal to terminal by a route that doesn't go through the battery, no non-conservative chemical work will be done on $q$ ! Path dependency is, then, starkly binary, in the case of the battery!

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  • $\begingroup$ Thank you for your answer. I understand that work is charge independent but is path-dependent but the ratio of work done over charge is constant. But why do batteries work like that? What is the reason $\epsilon$ is constant? $\endgroup$
    – Osmium
    Commented Dec 29, 2021 at 11:39
  • $\begingroup$ I'm not a chemist, but it's roughly like this: in a battery ions are released/discharged at the electrodes. The energy transferred per ion is the same for each ion, that is for the same quantity of charge that flows. $\endgroup$
    – Philip Wood
    Commented Dec 29, 2021 at 11:44
  • $\begingroup$ Consider two particles of the same charges crossing the battery by different paths. The work done on them will be in general different, but you said emf will be the same ie $\frac{w_1}{q}=\frac{w_2}{q}$. But then work done must be the same as well. But this in general is not true. That's a contradiction. There must be some mistake and that is that emf is path independent. $\endgroup$
    – Osmium
    Commented Dec 29, 2021 at 12:34
  • $\begingroup$ "The work done on them will be in general different" Why do you say this? $\endgroup$
    – Philip Wood
    Commented Dec 29, 2021 at 14:03
  • $\begingroup$ By that, I meant that the work on both particles does not need to be equal, it may be but it may not be. According to your statement, I showed that it follows that work done on such two particles will be the same whatever path they may take, always. Both statements are inconsistent and hence one or both is/are wrong. $\endgroup$
    – Osmium
    Commented Dec 29, 2021 at 16:30

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