Voltage between the positive terminal of a battery and a point in the air

[versine]

If I connect a voltmeter with the positive terminal of a battery and leave the other wire hanging in the air, it won't measure anything since it's not a closed circuit. But if there was a way to measure the voltage between those two points, what would the voltage be? Or is this a meaningless question?

 

[phinds]

You've answered your own question. You DID measure it. It's not a "closed circuit" so there's nothing to measure. Now if you want to get down to the really nitty gritty, it can be argued that there IS a circuit with a HUGE amount of resistance and some capacitance between the positive terminal and the other lead on the voltmeter, but we don't normally consider that as it's not particularly meaningful or helpful. I'm not even sure it can be measured with today's technology, although it might be.

Also, if you take that point of view then the concept of "open circuit" becomes problematic and that's one reason it's not helpful.

 

[tech99]

Between the terminals of a battery we have an electric field, and electric field lines are visualised running between the terminals (resembling the lines of magnetic force between the poles of a magnet). An exaggerated example of this is the field between the dome of a high voltage generator, such as a Van de Graaff machine, and its frame. It is possible to measure the voltage in such fields using an electrometer, which is a voltmeter having very high (for practical purposes infinite) resistance. If one side of the voltmeter (electrometer) is connected to the frame and the other moved towards the dome, a gradually increasing potential, or voltage, will be seen , showing the existence of a potential gradient. It is the potential gradient which would drive current if we provided a continuous circuit. An ordinary voltmeter and a battery will not show this.

 

[cmb]

If I connect a voltmeter with the positive terminal of a battery and leave the other wire hanging in the air, it won't measure anything since it's not a closed circuit. But if there was a way to measure the voltage between those two points, what would the voltage be? Or is this a meaningless question?

Working in very high voltages and resistances as I do, I am confident to say the issue here is your measurement equipment.

Consider that the air across the terminals is a very big resistor. Therefore, some distant point in the air will be half the voltage between the terminals.

If you have a 9V PPC sitting there on your desk, the voltage of the + will be at 4.5V wrt to the air at 20cm directly above it.

For the air directly between the terminals, it will be like a rheostat, a divided voltage.

That being said, two caveats;
1) You can't measure it because your meter will have an internal resistance of 1Mohm whereas the air at the end of the lead and back is >~10^12 that resistance (for several cm away). Your meter dominates the circuit from the terminal. You're effectively trying to measure the voltage drop across a 10^18Ohm resistor with a 1Mohm internal resistance DVM.
2) relative humidity will totally dominate this calculation. for high voltage calibrations, the air content has to be controlled for this reason.
3) because the resistance via the air is so high, like any poorly coupled circuit, the actual voltages wrt some midpoint in that air-voltage-divider can fluctuate easily if there is any disturbance directly at the potentials, for example if you breathe on one terminal or if some dust floats by, or some background EMI, can disturb the voltage.

Meanwhile, now consider you have a 90kV battery instead of 9V and ... you can see the resistance of the air begins to become a factor, which it does of course. It begins to break down due to the electric field it is bearing, and in turn the ionised products themselves become conductive.

With a high enough voltage battery and one end connected to ground, the other end of the potential will happily conduct itself through the resistive air to wherever that 'resistor' might go.

To deal with high voltages the solutions are either to evacuate the air out, removing that 'resistor', or put something else in its place that is an even higher resistor, such as a non conductive oil or, as in the case of most HV substation components, SF6 gas, although as it is a greenhouse gas there is probably some move to phase it out.

 

[Delta2]

This is a very good question actually. There is a voltage difference between the point in the air and the point in the pole of battery, however the voltmeter can't measure it.

That is because the voltmeter operates by measuring a current and then converting this current to voltage reading. But the current in this case is really very small, practically zero, because there is no closed loop (or actually as @phinds say, the closed loop is completed by air which has very high resistance, thus allowing really small current).

 

[rude man]

Voltage is the line integral of the electric field which, as others have pointed out, exists all around the battery, with field lines external to the battery pointing from anode to cathode.

If you somehow mapped the E field vector around the battery the voltage at any point outside the battery would be any line integral of E from the point to the anode, added to the battery emf if the cathode voltage is assumed zero.