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This question already has an answer here:

If electrons move from the negative side to the positive side in a battery (thus creating electricity when they flow), why does the same circuit with two separate batteries not create electricity? enter image description here

As you can see in the diagram, electrons should be moving from A to B because the wire is connected from A - side (high pressure) to B + side (low pressure). Why is there no electron flow?

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marked as duplicate by Chupacabras, Rev1.0, winny, pjc50, R Drast Dec 18 '17 at 10:45

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  • \$\begingroup\$ Also, that is not two batteries in a circuit. You haven't made a circuit of a running track until you go all the way round to where you began, in the same way it's not an electric circuit if it doesn't let the electrons close the loop. \$\endgroup\$ – Pete Kirkham Dec 18 '17 at 13:06
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Actually some current occurs when you connect the wires. It stops soon, maybe in few pico- or nanoseconds when the system has found a new balance.So, you cannot see anything, only some sensitive instruments could notice the change in the electromagnetic field or see the short light pulse. I even do not know if there exists light sensors that sensitive.

What balance? There really exists the electric field over the battery which pushes the electrons outwards from the minus pole and pulls them towards the plus pole. Only a medium where electrons can easily move is needed. If you connect the wire+led in your system, the electrons flow to the wire from battery A and some are taken off by battery B.

The basic fact: THE TOTAL AMOUNT OF CHARGE DOES NOT CHANGE. Battery A is soon positively charged and starts to pull electrons . Respectively B gets negative and starts to push electrons. The current weakens and stops when the generated opposite force compensates the pushing effect which is generated by the chemical reaction in the batteries.

In electric theory we have a concept named "capacitance". It depends on dimensions and materials. It tells how much charge a structure can take for a given voltage without having a fully closed conductive path for the current.

ADD due a comment: Inserting some load to both batteries by adding some conducting part between the poles of A and another between the poles of B does not make the current continuous in the original led+wire. The inserted loads do not remove nor add electrons in A nor B, they only make continuous chemical reaction possible inside the batteries.

Just for your information: Chemical reactions are electron transfers from a material to another. The new electron orbit configurations bind the atoms together in a new way and we see it as a formation or disbanding of chemical compounds. Some chemical reactions have so high tendency to happen without external energy input and change the eloctron orbit configuration so heavily that we can see the tencency as electric voltage.

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  • \$\begingroup\$ If that is the case if I add a discharging effect(lets say another led) connected individually to each battery(one led in each battery)....will that make the circuit work? \$\endgroup\$ – Gabriel Rodriguez Dec 18 '17 at 12:08
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Current flows in loops. A negative terminal is not a free flowing font of electrons. A postive terminal is not an unlimited sink for any number of electrons.

The battery creates voltage, or potential between its terminals - not between any terminal and some "aether" which connects everything together. (Air power does that, the "aether" is atmospheric air).

There is now a 2 x battery voltage potential between A+ and B-. It is patiently waiting for those two points to be conneccted, which would complete the loop.

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Current only flows when there is a potencial difference.The two batteries aren't connected so there is no potential difference. You cannot separate voltage and current like that, because the voltage makes the electons flow (current), when the circuit closed. And this circuit is not even closed.

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