Intuition about electromotive force (EMF)

[Kaushik]

My understanding of emf
Let us consider 2 parallel plates with charges (opp. but equal in magnitude) stored on it. When we connect both the plates from the outer side, the electrons from the lower potential (i.e., negatively charged plate) moves to the higher potential (positively charged plate) spontaneously. But this stops after some time as electrons get accumulated in the positive terminal and hence now the potential difference between the two plates is 0. So there is no current flow now. To ensure that the current flows, what we have to do is that, take the electrons from higher potential and move it to the lower potential by doing some work. We should do this continuously (from the start of the current flow) to ensure a steady current. This is done by the battery using some internal mechanism. So the work done by this non-electrostatic force to displace an electron from the higher potential plate to the lower potential plate is called the electromotive force (a misnomer).

Is this correct? If not, what is the mistake? Is there a better way to interpret this?

Also, When the current starts flowing, why does the terminal voltage decrease? But when there is no current flow, the emf is the same as the terminal voltage. Why?

 

[hutchphd]

In a battery sometimes the current is positive ions going the "other way" but your assessment is good enough. The change in EMF looks much like an internal series resistance and is caused by the chemistry not quite keeping pace with the demand for charge movement. Mostly this has to do with diffusion rates (which is why a cold battery is not as peppy) but I am not a chemist.

 

[rude man]

Your view of how emf is the "force" (actually field) that pushes electrons from the + plate to the - plate is essentially correct. The battery emf field does that. You can also see how this process will build up the capacitor's electrostatic field as more and more electrons are moved from + to - by the emf (the electrostatic field is building up). Equilibrium (no further current flow) is reached when the emf field in the battery cancels the electrostatic field in the capacitor. If you now disconnect the battery the charges stay put unless you connect the plates.

The terminal (capacitor) voltage decreases due to the internal resistance r in the battery. The battery emf is unchanged but the terminal voltage is reduced by the ir drop in the internal resistance. When current stops (i=0) there is no more ir drop and emf = terminal voltage.

I could expand upon this but maybe that's enough for this time around.

 

[vanhees71]

Despite the name an electromotive force is not a force in the modern sense. It's a relict from a time in the early 19th century, where "force" meant what we call "energy" today. E.g., one of the most important essays ever written at this time was Helmholtz' "Über die Erhaltung der Kraft" (literally translated: "On the conservation of force"). In modern terms what Helmholtz described was the general validity of the energy conservation law.

So the description in #1 is entirely correct (including the remark that EMF is a "misnomer" from the modern terminology; the only trouble is that nobody has invented a better name for "EMF" within 2 centuries!).

 

[Delta2]

On why the terminal voltage is reduced when we have current flow and internal resistance in the voltage source (e.g battery):
When there is no current the EMF equals the terminal voltage. When there is current with internal resistance r, the EMF has to overcome both the resistance r as well as the "resistance " of the electrostatic field (the field due to the terminal voltage $V_T$) that opposes the motion of electrons inside the source. So it has to be $$\mathcal{E}=V_T+Ir\text{ (1)}$$ and since the EMF $\mathcal{E}$ is fixed for a given source, equation (1) means that the terminal voltage $V_T$ has to become smaller, depending on the current $I$ and the internal resistance $r$.

 

[vanhees71]

What do you mean by "resistance of the electrostatic field"? I've never heard about such nasty creatures ;-)).

 

[Delta2]

What do you mean by "resistance of the electrostatic field"? I've never heard about such nasty creatures ;-)).

Because inside the source the electrostatic field does not accelerates the electrons but decelerates them. Therefore it behaves as some sort of drag or resistance.

Outside the source the electrostatic field accelerates the electrons.

 

[rude man]

Because inside the source the electrostatic field does not accelerates the electrons but decelerates them. Therefore it behaves as some sort of drag or resistance.

Outside the source the electrostatic field accelerates the electrons.

The "some sort of resistance" is Coulomb repulsion against further emf action. Balance is reached when the emf field is equal to in magnitude and opposite in direction from the electrostatic field.

The elctrostatic field is equal in magnitude and direction inside and outside the battery.

 

[hutchphd]

The "some sort of resistance" is Coulomb repulsion against further emf action. Balance is reached when the emf field is equal to in magnitude and opposite in direction from the electrostatic field.

Not even wrong...

 

[Delta2]

Balance is reached when the emf field is equal to in magnitude and opposite in direction from the electrostatic field.

That's when there is no ohmic resistance or impedance inside the source, or when there is no current. In the general case balance is reached when the emf field is equal in magnitude and opposite to the total "resistance" which is electrostatic field resistance +ohmic resistance+ reactance (in case of AC).

 

[rude man]

True, if there is current the net electrostatic field is reduced by the ir drop. As I said in my post 3.