# How does charge flowing between emf terminals reduce voltage difference?

I'm currently learning what electromotive force is and while reading my book's description of an ideal source of emf, I had difficulty understanding what these sentences mean:

The nonelectrostatic force maintains the potential difference between the terminals. If it were not present, charge would flow between the terminals until the potential difference was zero.

I don't quite get what it means by this. How does the nonelectrostatic force "maintain" the potential difference between the terminals? If it wasn't present and only the electric force remained, why would the potential difference decrease to zero?

The only thing I could think of was that the electric force from the E field inside the source of emf would still be exerting forces on charges from high potential to low potential, turning its electric potential energy into kinetic energy by doing work on the charges. I still don't know how that would reduce the voltage difference across both of its terminals to zero though.

Regarding the ideal source of emf, the book assumes there is a positive and a negative terminal with the positive terminal at a higher potential and the nonelectrostatic force going from low to high potential while the electric force goes from high to low, just in case that helps clear any ambiguity.

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.

• Thanks for taking the time to answer my question. I'd just like to clarify my understanding of electric potential which may be lacking and hence the cause of my confusion. Going back to the example with the spheres, the reason charge travels from the sphere with higher potential to the other is because the potential difference sets up an electric field in the wire that exerts an electrostatic force from higher potential to lower potential or did the electric field exist before the potential difference was created? I guess I'm asking why the charges move because of the potential difference. Oct 14, 2013 at 5:04
• A potential difference is exactly equivalent to an electric field, they are really two ways of looking at the same thing. You can think of the potential as being like the height of some landscape, the electric force as being like the force of gravity, and positive charge as being like water. If I have a potential difference, it's like having a hill, water will flow down the hill because of the force of gravity. [Negative charge flows up the hill so it's a little weirder]. So, in the two sphere example, one sphere is at a higher potential, meaning there's an electric field, meaning charge flows Oct 14, 2013 at 5:10
• OK, I should clarify that the electric field and the electric potential difference are proportional, not numerically equal. The [magnitude of] the average electric field between point 1 and point 2 is given in terms of the potential by $E=(V_2-V_1)/(x_2-x_1)$. Oct 14, 2013 at 5:12
• Based on the water analogy btw, you can think of a circuit as being like this escher painting: castletv.net/wp-content/forumuploads/shena/2012/12/…. In the painting it looks like the water is always flowing downhill, but in reality we know that if that was a true 3d setup then there must be a point in the 'water circuit' where the water is forced to flow uphill. Forcing the water to flow uphill is exactly what the battery does. Gravity can't be used to push the water uphill, just like the battery can't use electrostatic forces to maintain the potential difference. Oct 14, 2013 at 5:19
• That makes a lot more sense to me now, thank you. So going back to the quote in my question: if the nonelectrostatic force was not present in the ideal source of emf, the electrostatic force caused by the electric field would bring charges from the higher potential terminal to the lower potential terminal until the potential difference between the two would be zero. The nonelectrostatic force prevents this from happening and thus maintains the potential difference. If that's correct then I think I've got it! Thanks so much! I'd upvote you, but I need more rep first, sorry! Oct 14, 2013 at 20:36