I am learning about emf and I am using university physics of Hugh D. Young which states that when the emf source is not part of a circuit the non-electrostatic force of the source moves charge form the negative terminal to the positive terminal and that fact maintains the potential difference of the source.

When the emf source is not part of a circuit, the electrostatic force is equal to the non-electrostatic force so there is no charge movement.

How this is possible? What is the difference here?

  • $\begingroup$ Locally the electric field generated by emf is indistinguishable from an electrostatic electric field. How the charges in a circuit are separated is completely irrelevant for the analysis of the circuit. I don't know what else to answer, even though I feel that this is not what you are asking. Can you reformulate your question? $\endgroup$ – CuriousOne Dec 16 '14 at 4:19

I can't figure out your question, and wonder if it contains a typo.

However, consider a very simple system (simple in the sense that it's made of material you should know, and operates according to principles you've seen before). First use a conducting rod pointing in the x direction, and being moved at a steady speed in the y direction in a region with a uniform magnetic field in the z direction.

The magnetic force is our nonelectrostatic force. Due to the motion of the rod in the y direction the protons feel a force, but this just stresses the lattice and it strains the rod and it gets a bit bigger until there is an equal but opposite stress from the tension, so we can ignore the protons. Due to the motion of the rod in the y direction the conduction electrons feel a magnetic force so the charges pile up on one end, and leave an opposite imbalance on the other end. Eventually they pile up enough charge imbalance (like a capacitor) that the electrostatic force balances the nonelectrostatic force (the magnetic force in this case). During the time the rod charged its ends equally and oppositely the energy during the charging up came from the person pulling the rod. This is like a battery that is not hooked up to a circuit.

To hook it up to a circuit we can place a resistor parallel to the conducting rod but in in a region without a magnetic field (otherwise it'll be like another battery instead of like a resistor). We can then connect the rod and the resistor with two conducting rails that point in the y direction, but then the magnetic force just pushes them to one side of the rails, but doesn't drive them to one end or the other, so doesn't do anything important (the work function could be different on one side than the other). Now the charges won't pile up at the two ends of the rod, because each electron can just move away from the others down the rail, again there will be a very short period of time when the charges spread themselves out. But after they've done that, when the charge density isn't changing any more there are still electrostatic fields (the electric field caused by the location of charge density that isn't changing). And electrostatic fields have a curl of zero. So the forces are the electrostatic force which contributes no emf and the magnetic force which contributes only in the conducting rod.

So when the rod was by itself (no conducting rails, no resistor) it charged up its ends until the electrostatic force balanced the magnetic (nonelectrostatic) force. Later, when hooked up to circuit the electrostatic force smoothed the charge and current out to avoid charge build ups (unless there is a capacitor in the circuit), and so it wasn't fighting the magnetic force so the magnetic force can apply a force on the charges, the hand pulling the conducting rod can provide the energy and current can flow to the conductor where energy is lost as heat.

Other batteries work the same way, except they generate their charge buildup by different means (chemical, pressure, temperature gradients, etc.)

  • $\begingroup$ "Now the charges won't pile up at the two ends of the rod, because each electron can just move away from the others down the rail [...]" Is this really the case? I'd have thought it true only to the extent that the pd across the ends of the rod drops due to the addition of the external resistor. If the rod's 'internal' resistance were much lower than that of the external resistor, then the pile-up wouldn't be much diminished, surely. $\endgroup$ – Philip Wood Feb 1 '19 at 18:30

As far as I understand the question. The question is when EMF source is attached and detached.

EMF Source Attached.

Consider the EMF source to be a battery in a circuit.A battery maintains a potential difference between its ends and hence provides an electrostatic force for the charges in the circuit.

EMF Source Unattached.

When the EMF source (battery) is removed, there is no electric potential between any two points of the circuit and hence, there is no electrostatic force, therefore there is no movement of charge.


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