How would the surface charge density on the terminals of a new battery, say an AA, be determined? It is trivial to calculate the surface charge density of a fully charged flat plate capacitor of known plate spacing and voltage. But how would one do the same for the terminals of an AA battery?
 A: If you have an unloaded AA battery the electrodes have the   potential difference with the nominal voltage. To get the surface charges on the electrodes, you have to calculate the normal surface electric field on the electrodes by solving the Laplace equation for the electrostatic potential with the given boundary conditions, which depend on the shape of the battery and of the electrodes. From the gradient of the electrostatic potential you get the normal surface electric field on the electrodes from which follows the surface charge distribution on the electrodes of the battery. The problem is not easy to solve, and highly dependent on the geometry. It is similar to the case of a nonplanar charged capacitor
A: An AA battery has the ability to ’generate’ electrons using a chemical process. In an isolated battery these ions will charge up the battery’s connectors according to the one and a half volts that the chemistry provides and the capacitance of its configuration which will of course be minute. Therefore if the mechanical dimensions and separation of the conducting connectors are known then the charge stored can be calculated.
A: Perhaps, in an electrostatic sense, we can visualise our new isolated AA battery as comprising two point charges of plus and minus q coulombs separated by a distance of five centimetres?
Most importantly we are able to measure the voltage difference between the terminals as about 1.5V (usually 1.6
Computing the electric field between two charges gives a value of approximately a 3.2 x 10^13 x q newtons per coulomb. This was done by calculating the contributions from each terminal individually at the half-way point and adding them together.
We can also immediately calculate the electric field directly between the terminals because we know that the potential difference is 1.5V and the distance is 5 centimetres so the electric field must be 30 volts per metre.
Surely we can now equate the two and arrive at a charge of approximately a pico coulomb on each terminal?
Here is the answer why we see two AA batteries lying next to one another motionless: if one anode is near to the other's cathode, at say a millimetre distance, then the Coulomb Force is simply 9 10^9 x 10^-12 x 10^-12 /(10^-3)^2 N.
If my arithmetic is correct then at a millimetre separation there is an attractive force of approximately one hundredth of a micro newton. The weight of a battery is about 23g. Using Newton's second Law then at a distance of one millimetre the batteries will be accelerated towards each other at approximately 0.4 10^-6 m/s/s. So the batteries will be attracted towards each other IF THEY ARE ALREADY VERY CLOSE, but over a period of some minutes which would be impossible to spot other than under experimental conditions.
But I haven't taken into account friction. This would most likely overwhelm the minute attractive force and the batteries would remain in situ.
All the above is imagining that the battery comprises two equal and opposite point charges of a pico coulomb separated by 5cm and with voltage difference of 1.5V. The actual battery construction is absolutely nothing like that in terms of internal charge separation so the actual charges on the battery terminals will be considerably different and an equivalent point charge setup will require a great deal of thought!
