If electrons move from the negative side to the positive side in a battery(thus creating electricity when they flow), why doing the same circuit with two separate batteries doesn't 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 in A - side(high pressure) to B + side(low pressure)...so, they should flow...why they aren't?

I've heard some people say that it is because of the balance(not enough to keep the constant flow). But then shouldn't it work if battery A has less charge than B?


The short answer is that you need a complete circuit for a battery to work. However, I find a longer answer to be very useful to help with understanding.

If you have a circuit, you are solving a problem in what we call "electrodynamics." Electrodynamics is studying what happens when electrons are moving. In a circuit, electrons are constantly circling around the loop (powered by the Electromotive Force (EMF) of the chemical reactions in the battery). These moving electrons can do work, such as lighting an LED (which decreases the energy in the electrons and emits that energy as photons of light).

If you have no movement of electrons, you have an "eletrostatics" problem. We see electrostatics all the time, like when you scuff your feet on the carpet and then get shocked on a doorknob. Electrostatics is interested in what things look like when the electrons finish moving. Without anything like a chemical reaction to keep the electrons moving, they do eventually settle into a lowest energy arrangement.

The setup you describe sits in the fuzzy region between the two. If you let it sit, it has to become an electrostatics problem, because there's no loop for the electrons to flow around in. They have to eventually find a lowest energy setup. However, your intuition is correct that the electrons are going to have to flow a bit in order to arrive at this arrangement.

We're going to have to look at what happens when you hook all of these elements together. But first, a warning: the electrostatics side of this problem is going to be rather boring. Typical electrostatics problems involve high voltages. 200V to 20kV are not uncommon in electrostatics. Our little batteries can pump out 1.5V, so expect the final electrostatics solution to this problem to be rather uninteresting. It's the transients that are interesting.

Before we hooked everything together, the chemical reactions in each battery have pushed more electrons towards the negative side of the battery and fewer towards the positive side (there's more positive ions on the positive side). This chemical reaction will continue to push the electrons like this, but as it does, we start to build up a negative charge at the negative side (and a positive charge at the positive side). The electrons bundled up on the negative side start to repel, and this repelling actually fights against the chemical reaction, making it less and less likely to occur. Eventually the whole system reaches an equilibrium, at the point where the voltage difference between positive and negative sides is 1.5V (for normal AA batteries).

Now you bring the system together. Your intuition is actually going to be correct at first. We've got a bunch of electrons at the bottom of the bottom battery, and a bunch of positive ions at the top of the top battery, and a wire between them. Electrons will indeed flow from bottom to top, and they'll do some work as they pass through the LED, generating some light. However, we're going to find this is a VERY small amount of light, because this process happens really fast.

Now that there are fewer electrons at the bottom of the bottom battery, the charge difference between positive and negative sides of the battery is smaller, so the chemical reaction isn't inhibited as much. It starts reacting, pumping more electrons across to the negative side. The same occurs in the top battery, because there's not as many positive charges at the top of the top battery.

But after a very short time, this process slows. Soon we have so many positive ions at the top of the bottom battery, and so many electrons trapped at the bottom of the top battery that, even with the wire connecting them, there's still enough electrostatic charge within each battery to prevent the chemical reaction from going forward any further.

So now we arrive at the electrostatics part of the problem. There's 3V of potential between the positive terminal of the bottom battery (chock full of positive ions) and the negative terminal of the bottom battery (full of electrons). There's enough charge there such that neither battery's chemical equations "want" to move forward. Thus, there's no electrons flowing, no electrons doing work, so the LED stays unlit.

This process happens fast. It's far faster than you can see. We can add some elements like capacitors to slow it down, but with these batteries, you'll never actually notice this effect occurs. However, if you work with static sensitive parts, you'll learn this process can occur in an instant and burn your parts. A measly 3V of static potential like we have here won't damage anything. However, if you aren't careful, things like wool sweaters can easily build up very large static charges, on the order of thousands of volts. If you are charged up like this and touch the circuit, it's exactly like your batteries here, only on a much larger scale. Many more electrons rush out of your body, seeking to equalize these potentials. All of them do "work," which in the case of static sensitive parts can include things like burning through precious oxide layers and rendering them inoperable!

  • $\begingroup$ Thanks for the thorough explanation..when you say: "Soon we have so many positive ions at the top of the bottom battery, and so many electrons trapped at the bottom of the top battery that"...wouldn't it work then if battery A is less charged then Battery B?....BTW: I thought that internally in battery the positive side was totally separated(isolated) from the negative side(like in two separate atoms containers), aren't they? $\endgroup$ Dec 18 '17 at 15:10
  • $\begingroup$ @GabrielRodriguez If battery A was charged less than battery B, there would just be a different balance of charges before the EMF from the chemical reactions was balanced out. As for the internal battery components, the negative and positive sides are not connected, as in there's no wire between them, but there is an electrolyte which is touching both sides. Wikipedia would be a good place to start for understanding how batteries work on the inside. $\endgroup$
    – Cort Ammon
    Dec 18 '17 at 15:16

A battery in a complete circuit will do work on electrons, pushing them round the circuit. Electrons leaving the negative terminal of the battery are replenished by electrons entering the battery at the positive terminal. But if the battery is not connected to anything, some of the electrons inside it will pile up on its negative terminal, and there will be a deficit on its positive terminal. But these charges won't grow and grow, because like charges repel each other, so there's a limit to the amount of charge that can pile up. [The potential difference between the terminals due to the charges on the terminals can't exceed the emf of the cell, the work it can do per unit charge.]

In the set-up you've drawn, when the LED is first connected, some electrons will flow through it, from the negative terminal of A to the positive terminal of B. Not enough will flow to light the LED. This is because battery A as a whole will have acquired a positive charge (having lost electrons) and battery B as a whole will have acquired a negative charge. These acquired charges will prevent further electrons flowing through the LED.

  • $\begingroup$ If that is the case, if I add a discharging effect(lets say another led) connected individually(isolated from the circuit) to each battery(one led in each battery).......will that make the circuit work? $\endgroup$ Dec 18 '17 at 14:38
  • $\begingroup$ I'm not sure exactly how you propose to connect these extra leds, but there'll be no sustained current unless the batteries are in complete circuits! $\endgroup$ Dec 18 '17 at 15:10

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