I actually have three related questions: An open circuit chemical cell separates charges creating a surplus of electrons on its negative terminus and a shortage of electrons on its positive terminus. The separation of charges is conducted by chemical reactions. As electrons accumulate on the negative electrode, they repulse each other, so it takes more work to put any additional electrons in the negative electrode, so eventually, after a certain potential difference is achieved, the chemical reactions are inhibited and stopped.

1) If we connect a long wire to the negative terminus, the accumulated electrons will evenly distribute themselves along the wire, because of their mutual repulsion. Thus, the charge density at the negative electrode will decrease. Will this cause the battery to relaunch the chemical reactions and put more electrons in the negative electrode?

Let's say we connected two long wires to a chemical cell, one to it's negative electrode, another to it's positive electrode (the wires do not connect to each other). The negative wire will thus acquire some extra electrons, the positive wire will have a shortage of electrons.

2) If now we disconnect the wires from the battery, will they remain charged? And how is this charge mathematically related to the emf of the battery?

If we connect the wires to the battery as described above, the potential difference between their termini will be about the same as the emf of the battery.

3) If we disconnect the wires from the battery and they remain charged, what will be the potential difference between the wires?

Thank you.


3 Answers 3


Suppose we take some object, for example a conducting sphere, and start with it at the same electrical potential as its surroundings. Now we add one electron the sphere, and because the sphere now has a higher negative charge that its surroundings the potential of the sphere will be slightly lower (i.e. more negative) than its surroundings. Adding the electron has created a potential difference. Now add a second electron and the potential difference gets bigger. Add more and more electrons and the potential difference keeps getting bigger.

A side note: this is basically how a Van de Graaff generator works. Charge is added to the sphere at the top by mechnical means, and this can create a big enough potential difference to generate some impressive sparks.

Anyhow, if we transfer a charge $Q$ to our metal sphere the resulting voltage change is given by:

$$ V = \frac{Q}{C} \tag{1} $$

where the constant $C$ is called the capacitance of the sphere (strictly speaking it's the self-capacitance). For spheres we can work equation for the out the capacitance fairly easily, and in fact it's given by:

$$ C = 4\pi\varepsilon_0 r $$

where $r$ is the radius of the sphere. For objects with different shapes the equation for the capacitance will be different, but the key point is that any object of any size and shape has some capacitance and we can use equation (1) to work how what voltage difference is created when we add charge to it.

The point of all this is that the ends of your battery also have a capacitance. Suppose the chemical reaction in the battery transfers a charge $Q$:


The if the capacitance of the anode is $C_-$ and the cathode $C_+$, the voltage change due to transfer the charge will be give by equation (1) as shown on the diagram. The total voltage will be:

$$ V = V_+ - V_- = \frac{2Q}{C_b} $$

where I've assumed the capacitances of both ends are the same and I've used $C_b$ for them both. What actually happens in a battery is that the reaction runs and transfers charge until the voltage $V$ builds up to the battery voltage. At that point no more charge can be transferred and the reaction stops.

This probably all seems a bit long winded, but having gone gone through all this we can answer your questions really easily:

  1. when you attach a long wire to the ends of the battery you will increase the capacitance. The battery voltage is constant, and we can rearrange equation (1) to give:

$$ Q_T = \tfrac{1}{2} C_T V \tag{2} $$

where $C_T$ is the new bigger total capacitance with the wire attached. So when you increase the capacitance, $C_T$, you increase the charge $Q_T$. That means the reaction in the battery will restart and transfer more electrons until the total charge rises to $C_T$.

  1. the electrons spread out across the ends of the battery and the wire. If we call the capacitance of just the wire $C_w$ and the capacitance of the end of the battery $C_b$, then the total capacitance is $C_T = C_w + C_b $. If we put this into equation (2) we get:

$$\begin{align} Q_T &= \tfrac{1}{2} (C_w + C_b) V \\ &= \tfrac{1}{2} C_wV + \tfrac{1}{2} C_bV \\ &= Q_w + Q_b \tag{3} \end{align}$$

where $Q_w = \tfrac{1}{2} C_wV$ is the charge on the wire and $Q = \tfrac{1}{2} C_bV$ is the charge on the battery. So if you disconnect the wire it keeps a charge $Q_w$. The size of the charge is given by $Q_w = \tfrac{1}{2} C_wV$.

  1. The potential difference between the wires is just the battery voltage $V$. If you connected them together with some suitable voltmeter, that meter would show a voltage of $V$.

In practice the capacitance of a piece of wire is going to be very small, and the charge you'd build up on the wires will be tiny. Nevertheless, the capacitance will be greater than zero so there will be some charge.

  • $\begingroup$ Wow! Thanks a lot. It will take a while for me to absorb this. $\endgroup$ Commented Jan 29, 2015 at 17:44
  • $\begingroup$ Thank you, John Rennie, for such a detailed answer. There is one thing about it that still confuses me. Let's say we attached an open circuit wire to the negative electrode of a battery. Let it all be in the vacuum. The battery puts some extra electrons into the wire, making it negatively charged. We disconnect the wire from the battery. Now we have a wire with extra electrons in it. We take another wire, and connect it to the negative electrode of the battery. Disconnect it. Connect another wire, and so on. $\endgroup$ Commented Jan 29, 2015 at 18:39
  • $\begingroup$ Each wire will take the same amounts of electrons from the battery, leaving the battery as a whole positively charged. Thus over time, the positive charge on the battery will increase, making it impossible for the electrons to flow from the battery into a wire. So, this does not seem to work at all unless there is a closed circuit. What do you think? $\endgroup$ Commented Jan 29, 2015 at 18:40
  • $\begingroup$ @SantaClaus: the chemical reaction that takes place in a battery is basically an electron pump. It does not create electrons. It just pumps them from the cathode to the anode. If you connect identical pieces of wire to the two terminals the battery will pump electrons from one wire to the other, leaving one wire positively charged and the other with an equal and opposite negative charge. If you only connect one wire (to the anode) then the battery can only pump a very limited number of electrons onto it because it has nowhere to get the electrons from. $\endgroup$ Commented Jan 29, 2015 at 20:11
  • $\begingroup$ To go back to my answer, if the wire is only on the anode then the capacitance $C_- >> C_+$. That means the voltage $V_+ = Q_+/C_+$ builds up very quickly and stops electrons being pumped. $\endgroup$ Commented Jan 29, 2015 at 20:12

Regarding Point 1:
You have to consider the electrical field in the battery, too. As soon as you remove electrons, you get "holes" at the positive end of the battery. The more electrons you remove, the bigger gets this field, stopping electrons from further going out of the battery. This would only work if you remove the positive charges (aka "holes"), too. Same goes if you connect an additional wire to the positive pole of the battery: It gets more positive charges, but the internal electrical field stops them soon.
Regarding 2:
If you have wires in the air which are charged they will discharge quite fast because of the "free" electricity in the air (or, if you touch them, then via your body). Thus rendering 3 not possible (except free floating in the vacuum)...

  • $\begingroup$ Thank you, arc_lupus, for the answer. I see your point, but I am still in doubt. Would you care to elaborate as to what are the exact answers to my three questions? Let it all be in vacuum. $\endgroup$ Commented Jan 29, 2015 at 17:34

1) Attaching a wire to the electrode will cause more electrons to be generated in the negative side of the battery in order to load up the wire with some of them. They will not generally distribute themselves evenly, though. They will distribute themselves so as to create an equal potential (voltage) at every point along the wire. How many electrons need to be at each point will depend on how far that part of the wire is from the positive electrode and any wires attached to the positive electrode. The further a piece of wire on the negative electrode is from anything connected to the positive electrode, the fewer extra electrons will be on that part of the wire.

2) If we attach a wire to each electrode of the battery, then the wires will be charged with approximately equal and opposite charges. If the wires are moved towards each other, more charges, more electrons in the negative wire and more "holes" in the positive wire, will be on them. If they are moved away from each other, the amount of charge goes down. In terms of electronic ciruits the two wires have a capacitance C which is higher when the wires are closer together. The net charge on each wire is $+/- q$ where $q=CV$ where V is the voltage of the battery.

3) If the wires are disconnected from the battery but in that process do not touch a grounded conductor, for example they are disconnected using plastic wire cutters by someone wearing rubber gloves, then the charges on the two wires will stay there, stuck until the wires are touched by a conductor. If the wires are cut from the battery but otherwise left in place, the voltage between the wires will be equal to the voltage of the battery. If the cut wires are moved closer to each other, the voltage will go down. If the cut wires with the charge stuck on them are moved further apart, the voltage between them will go up. Note moving two oppositely charged wires apart from each other costs energy, they attract each other and you have to pull them apart, and the energy you put into this effort is what supplies the energy to raise the voltage between the wires. What is happening here is explained by $q=CV$ except transformed to $V=q/C$. $q$ is not changing as we move the wires closer and further from each other wearing rubber gloves. But since C is changing, the voltage on the wires will be changing.

  • $\begingroup$ Thank you for your answer. Now I see that your answer was correct as well. I wish I could vote it up, but I do not have enough reputation for this. $\endgroup$ Commented Jan 30, 2015 at 16:38

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