Taking the battery as an example. Intuitively, I understand that without the chemical forces that continuously maintains a voltage difference between the two terminals, the current will stop because, eventually, the two terminals of the battery will obtain the same voltage.

However, most textbooks I have read emphasize that the source of an EMF must be non-electrostatic (or non-conservative) in origin. From my understanding, as stated above, I don't see a connection between the chemical forces that maintain a voltage difference between two battery terminals and that they must be non-electrostatic (or non-conservative). It seems to me that the mechanism of maintaining a voltage difference (in this case, the chemical forces) still works just fine without the requirement of the forces being non-conservative.

My question is: Why is the fact that the source of an EMF being non-conservative important? I'd prefer an explanation in the context of my understanding of EMF, which is stated in the first paragraph. Also, please limit the explanation to classical EM only.

If my understanding is missing something, which might have rendered my question to appear "nonsense", please help me correct my misconception(s).


1 Answer 1


There is a lot of interesting stuff going on behind your textbooks' emphasis.

A conservative force is one which can be described as the gradient of a potential function. We speak of gravitational and electrostatic potential because gravity and static electricity produce conservative forces. This does mean that the total change in potential around any wire loop must be zero, but just because the total work done on a test charge going around a loop is zero does not mean that all of the forces on the test charge are conservative.

Within the battery, charged ions are flowing from a terminal where they have low potential energy to one where they have high potential energy. So they must be being pushed by some other force, one which is not associated with a potential--a "non-conservative force". This is the observation which your textbooks are emphasizing, and it is a fact that can be fairly hard to wrap your brain around at first. I know it was for me.

The non-conservative force at work inside an acid battery is the random jostling of the liquid molecules surrounding the charged ions. The positive ions are being electrostatically pushed toward the cathode, but for chemical reasons they are being produced faster than they are being absorbed at the cathode. At the anode, they are being consumed faster than produced, so the net effect of the jostling is to move positive ions "uphill" to the anode. Analogously, negative ions are jostled "uphill" to the cathode. So, in a very non-literal sense, the second law of thermodynamics is creating the EMF.

Pretty crazy, right?

PS. The resistance of the wire is a second non-conservative force, which exactly undoes the work done by the random jostling in the battery. This makes the total work done on a test charge going around the circuit zero, even though there are non-conservative forces at work in addition to the conservative electrostatic force.

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    $\begingroup$ Thank you. As far as I understand, the non-conservative chemical force is confined to the interior of the battery only. However, my textbook also mentions that the chemical force inside the battery must exactly balance the electric force in order for charges to be pushed uphill. Thus, if the chemical force has already been balanced by the electric force, how could it still provide energy for charges to overcome the resistance of the external wire? $\endgroup$ Feb 20, 2018 at 6:16
  • $\begingroup$ Good question. Each non-conservative force is directly balanced by one side of the electrostatic potential hill, but the two sides of the potential hill have the same magnitude because the loop is closed. Thus the two non-conservative forces are balancing each other indirectly. $\endgroup$ Feb 20, 2018 at 6:27
  • $\begingroup$ Thank you. If I understand you correctly, the non-conservative chemical force overcomes the electric force inside the battery to push charges uphill. Once charges are at a higher potential, they overcome the wire resistance and go downhill to a lower potential. Is this correct according to what you explained? $\endgroup$ Feb 20, 2018 at 16:38
  • $\begingroup$ That is correct. $\endgroup$ Feb 20, 2018 at 23:42
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    $\begingroup$ @The Oblivious Your hypothetical situation would not work as you suppose, because any conservative force does zero net work around a loop, by definition. It is not possible, in an isolated system, for any combination of static potential functions to drive current through a resistive loop. An electric field might be able to replace a battery in driving charge from on battery terminal to the other, but in doing so it would eliminate the potential difference which was driving current through the wire back to the start of the loop. No current would flow and it would not act as an EMF. $\endgroup$ Feb 22, 2018 at 19:04

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