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Apr1-11, 01:40 AM
P: 431
So what would cause the potential of a 12 volt battery connected to that open conductor to go from 12V to 0?
That would be because there is an electric field inside the battery and outside it--the ∫E.dl around a loop of which will be zero.

Why would the two fields not simply add together? Why would they add to 0?
Yes they would. But I used the battery as an example. The original question did not have it. I used the example of an isolated battery to bring out a certain point.

The electric field produced by the battery and the induced field do not add to zero but they do produce a static electric field by causing charges to be accumulated at the ends of the ring.That is like a capacitor. That field stops further accumulation of charges and gives us a finite potential.

Not quite. The difference is that for a circuit containing a battery and no induced field ie. over the area enclosed by the circuit. That is the fundamental difference.
The ∫E.dl is -dφ/dt but that happens over a cycle. The loop is incomplete and hence no area or flux can be associated with the integral.

I am sorry but I have trouble putting my points across.