I have read almost all answers on PSE and ESE but none of them are to the point or maybe I am not understanding.
I have described the setup in the title , I just want to know what will happen , will there be a momentary current / nothing happens.
Correct me if I am wrong but batteries have surplus of electrons on one terminal and deficit on other so they are at non zero potential with respect to infinity , so shouldn't there be a current , if not then atleast a force on free electrons of wire because wire is connected to + and - terminals on either end
I have read people answers where they say battery potential is not absolute to disgrace the above argument , therefore I've specifically told identical batteries so indeed there will be a potential difference across the wire.

from my other questions I have been able to gather that a piece of wire connected to a single terminal does get charged because that terminal of battery want to make it equipotential with its terminal , this is by considering the wire as capacitor plate.

so in my current question the opposite terminals will try to charge it differently so current flows?
I'm really confused please clear this once and for all with in depth explanation of all arguments and counter-arguments


In reality there are always parasitic capacitors between any 2 points in the circuit (just like any wire is an inductor). But their value is extremely small and typically is ignored until you need to consider GHz or nanosecond range.

So current will flow until all these parasitic capacitors are equialized, but it will take fraction of nanosecond due to their extremely small capacitance (pF range). This will be quite hard to detect.

Wires also do have some parasitic capacitance, so when you connect wire to 1 terminal of the battery - current also flows momentarily to charge this capacitor to match battery potential.

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  • $\begingroup$ so if I connect only one plate of capacitor to one terminal will it get same charge as it would have got had there been its parter plate ? but how could this be possible because battery Dr was electron from plate and gives to other keeping its terminal potential same in the process , here there is no partner plate to draw the electrons from so ? $\endgroup$ – Aditya Prakash May 23 '19 at 23:13
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    $\begingroup$ Any 2 non-connected metal objects is a capacitor. So there is always a second plate. 2 scews that lie on our desk is an air-spaced capacitor. In your example parasitic capacitor is between capacitor case/lead and other terminal of the battery which is quite far away. It will have very small capacitance (<1pF). $\endgroup$ – BarsMonster May 23 '19 at 23:22
  • $\begingroup$ that is what I was confused about , the second plate being the terminal itself ? it should charge up , even a little won't terminal voltage change now as opposed to a real conductor being second plate where the terminal had a source to draw electron from $\endgroup$ – Aditya Prakash May 24 '19 at 3:29

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