# Electric potential difference between a battery's + terminal and the ground

cianfa72
Hi,

I've a question about electricity in the following scenario: consider an accumulator (e.g. a 9V battery) and an analog/digital voltmeter having a probe connected to the accumulator + clamp and the other to the ground (for instance connecting it to a metal rod stuck in the ground).

Do you think there will exist a electric potential difference as measured by the voltmeter ?

Thanks.

Periwinkle

Homework Helper
No, it will measure noise

davenn
Gold Member
Hi,

I've a question about electricity in the following scenario: consider an accumulator (e.g. a 9V battery) and an analog/digital voltmeter having a probe connected to the accumulator + clamp and the other to the ground (for instance connecting it to a metal rod stuck in the ground).

Do you think there will exist a electric potential difference as measured by the voltmeter ?

Thanks.
Unless there is an electrical path through the meter and to both battery terminals, there will be no measurable PD. If you could clarify the actual circuit that you are discussing by means of a diagram, it could help but your actual wording is not absolutely clear. (i.e. are there any more connections / wires involved?)

Periwinkle
I think that when the battery was manufactured, it was made of materials and equipment that was at ground potential 0. As a result of the separation of the charges, the + pole has a higher potential while the negative pole has a lower potential than the Earth.

davenn
Homework Helper
No, all the battery provides is a potential difference between the poles. You can add potential differences by connecting cells in series, as in e.g. a car battery.

Mark Harder
Periwinkle
No, all the battery provides is a potential difference between the poles. You can add potential differences by connecting cells in series, as in e.g. a car battery.

Objects have some potential compared to Earth. Everything has some potential compared to Earth, such as the Moon and Mars.

davenn
cianfa72
If you could clarify the actual circuit that you are discussing by means of a diagram, it could help but your actual wording is not absolutely clear. (i.e. are there any more connections / wires involved?)
Simple schema, there is no close electrical path: voltmeter probes are connected one to the + battery clamp and the other one to the ground.

My concern was related to the fact that, as others said, thanks to the separation of the charges the + pole should assume a non-zero electric potential respect to the Earth

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Homework Helper
Objects have some potential compared to Earth. Everything has some potential compared to Earth, such as the Moon and Mars.
OP asks about a voltmeter, not an electroscope. Any charge leaks away long before you can measure a voltage

rbelli1
cianfa72
OP asks about a voltmeter, not an electroscope. Any charge leaks away long before you can measure a voltage
Can you elaborate it please ? As far as I can understand when charges leaves the battery pole reaching the ground through the voltmeter actually they are replenished by the battery electromotive force (emf) I guess..

Homework Helper
Everything has some potential compared to Earth
which refers to the electrostatic potential you get when you load some object with charge, e.g. by rubbing it. It has nothing to do with the chemical potential difference as generated by a battery cell.

In your case, what would you expect if you connect the ##\ +\ ## from one battery in your scenario with the ##\ -\ ## from another ? Do you really expect both batteries to empty themselves, completely contrary to what happens in reality (namely: nothing) ?

A multimeter measures potential difference by the flow of electrons through it. A battery has no net free charges so there is nothing for the multimeter to use. The potential difference of a battery is produced only when the a chemical reaction occurs in the battery. This can only occur when a conductive path is created between its electrodes which allow electrons to flow from the cathode where a material gives up electrons to the anode where an ion accepts the electrons completing the reaction. So a multimeter will only measure a potential difference between its electrodes.

Klystron, thomasj, sysprog and 3 others
Periwinkle
which refers to the electrostatic potential you get when you load some object with charge, e.g. by rubbing it. It has nothing to do with the chemical potential difference as generated by a battery cell.

I just tried to think logically. I think there is only one kind of electricity. I also believe that I may not be right, I just don't know the reason. I don't understand what is the chemical potential?

Mentor
I just tried to think logically. I think there is only one kind of electricity. I also believe that I may not be right, I just don't know the reason.
But it can be very confusing for folks like the OP when you make replies that are so incorrect.
I don't understand what is the chemical potential?
https://en.wikipedia.org/wiki/Electric_battery

Gold Member
Simple schema, there is no close electrical path: voltmeter probes are connected one to the + battery clamp and the other one to the ground.

My concern was related to the fact that, as others said, thanks to the separation of the charges the + pole should assume a non-zero electric potential respect to the Earth
The PD between the battery and the Earth is quite arbitrary, depending on the carpet you were walking on just before the measurement, for instance. A sensitive (and isolated) enough instrument would yield two different values for the Potential on the + and - terminals because the Electric Field Vectors do actually add vectorially.
If, instead of the battery being totally isolated, you put it on a sheet of graphite paper, with one edge held at Earth Potential there would be a small flow of charge from + to - terminals and the 'half way' point between the terminals would be at the same potential as the Earth connection but the + terminal would be +4.5V relative to Earth and the - terminal would be at -4.5 V.

cianfa72
If, instead of the battery being totally isolated, you put it on a sheet of graphite paper, with one edge held at Earth Potential there would be a small flow of charge from + to - terminals and the 'half way' point between the terminals would be at the same potential as the Earth connection but the + terminal would be +4.5V relative to Earth and the - terminal would be at -4.5 V.
Not sure to understand your experiment involving the battery and the sheet of graphite paper...could you help me ?

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Gold Member
Not sure to understand your experiment involving the battery and the sheet of graphite paper...could you help me ?
Sorry - a bit shorthand there. I mean you connect the two battery contacts to points on the paper (the battery will pass current then and you don't need to consider the idealised and indeterminate 'floating in air situation'. If you connect the meter between each terminal and the mid point, you will get + or - 4.5V and 9V across the two contacts and that goes for connection to the edge of the paper (many times the spacing between the terminals. If the paper is connected to Earth we can call that 0V 'absolute' and you will have your two different PDs to Ground.
I am basically suggesting repeating the thought experiment in a way that will actually give you an answer without needing a fancy Voltmeter with infinite input resistance which, as @BvU suggests, will just pick up random E fields.
If you want to see an interesting picture of Potentials, connect the meter to one terminal and put the other probe somewhere on the paper. Then move the probe over the paper to follow the same PD value. These will be the contours of equal PD.

Periwinkle
which refers to the electrostatic potential you get when you load some object with charge, e.g. by rubbing it. It has nothing to do with the chemical potential difference as generated by a battery cell.

In your case, what would you expect if you connect the ##\ +\ ## from one battery in your scenario with the ##\ -\ ## from another ? Do you really expect both batteries to empty themselves, completely contrary to what happens in reality (namely: nothing) ?

I was undoubtedly wrong, but my idea was just a thought experiment, and anyway, I hadn't read such things for a long time.

However, with that, an electrostatic potential-difference exists and a distinct chemical potential difference, I totally disagree.

The only thing here is that as long as the two metals are not connected by a wire, the chemical processes that produce the potential difference will not start. However, there is only one kind of potential difference.

Gold Member
Simple schema, there is no close electrical path: voltmeter probes are connected one to the + battery clamp and the other one to the ground.

As was said much earlier in the thread ...
so there will be no potential difference, other than maybe some noise that the meter picks up

Gold Member
However, with that, an electrostatic potential-difference exists and a distinct chemical potential difference, I totally disagree.

disagree with what ??
As @BvU said, you are confusing the OP with info that is totally irrelevant to the OP question asked

cianfa72
If you connect the meter between each terminal and the mid point, you will get + or - 4.5V and 9V across the two contacts and that goes for connection to the edge of the paper (many times the spacing between the terminals. If the paper is connected to Earth we can call that 0V 'absolute' and you will have your two different PDs to Ground.
If I got correctly, the measured PD between each point where the battery poles are connected on the paper and the edge of the paper itself (many times away the spacing between the terminals as connected to the paper) will be the same as the PDs measured between each of them and the mid point between them on the paper (as far as I can understand you are basically building a "distributed potential divider" using a graphite sheet). Then, if you connect the paper edge to the Earth (ground) we can assume midpoint potential is actually 0V.

does it make sense ?

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Gold Member
"distributed potential divider" using a graphite sheet)
You've got it!
we can assume midpoint potential is actually 0V.
Yes. There will be a 0V line at right angles to the line of centres which goes from the mid point to the edge. The equipotential lines will be tightest between the two terminals and more spread out, further away. And that is a direct model of what you would see in the space around a battery, isolated in space. I think it's a good alternative way into the problem.

cianfa72
And that is a direct model of what you would see in the space around a battery, isolated in space. I think it's a good alternative way into the problem.
Yes, it should be the characteristic layout of an electric dipole in terms of equipotential lines.

Coming back to the original question
The potential difference of a battery is produced only when the a chemical reaction occurs in the battery. This can only occur when a conductive path is created between its electrodes which allow electrons to flow from the cathode where a material gives up electrons to the anode where an ion accepts the electrons completing the reaction.
Thus, in order to get a potential difference between the battery poles is absolutely required to have a conductive path in place between them ?

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Gold Member
Thus, in order to get a potential difference between the battery poles is absolutely required to have a conductive path in place between them ?
Not 'as such'. At some stage, charges needed to be moved around, to set up that Potential distribution. If you want to do that as a 'once-off', you can move charged balls on sticks but, to maintain that particular Potential setup, it is more reliable to do it with a 3D Potential Divider, which has a continual current - albeit small.
It's a bit like a distribution of Masses having a Gravitational Potential distribution. That involves moving the masses in the first place. If those masses were continually evaporating and their content was being blown away in the wind, you would need to 'pump' more mass to maintain the situation.

You can determine the potential difference of a battery using a potentiometer (not the common voltage divider or pot). The instrument does not draw from or supply current to the battery, so no chemical reaction can take place.

cianfa72
You can determine the potential difference of a battery using a potentiometer (not the common voltage divider or pot). The instrument does not draw from or supply current to the battery, so no chemical reaction can take place.
I think my basic concern is this: to build up a charge separation inside the battery (that ultimately is responsible for the potential voltage difference between electrodes by means of internal chemical reaction) is a conductive path between electrodes required or not ?

I think my basic concern is this: to build up a charge separation inside the battery (that ultimately is responsible for the potential voltage difference between electrodes...

A charge separation is not responsible for the potential difference. The potential difference is inherent in the chemical entities that are interacting. As best as I can explain it the charges are transferred because one element of the chemical process has a higher attraction for electrons than the other. The valence electrons of one element is more loosely bound than the other. Take the simple Daniell cell that uses Zn and Cu as the interacting elements. On discharge Elemental Zn more easily give up it valence electrons to Cu ions resulting in a lower overall energy of the pair. If you put metallic Zn in contact with Cu ions as in a CuSO4 solution the Zn will dissolve loosing its valence electrons to the Cu ions. In the Daniell cell the Zn is isolated from the Cu ions. When the Zn cathode is connected to the Cu anode the affinity of Cu for the Zn electrons is transferred via the conductor because of the potential difference inherent in the pair.

mselectromagnetic and thomasj
Gold Member
A charge separation is not responsible for the potential difference.
+1 (I think.) It's a bit like chicken and egg though. The electronic structure of the material in the electrodes is what it is because the charges take up a minimum energy state so which is the cause and which is the effect, here? When the electrolyte comes into contact with the electrode surfaces, charges 'fall' into the electrodes and there has to be a Potential Energy cause.
At some stage you need to draw a line around, say, the (excess) charges in the battery terminals and start your explanation about what goes on outside from there. Problem is that however you move charges (electrochemical reaction or 'on sticks', it all relies on the EM set up in the thing that causes the charge movement.)
Scope here for a lot of pointless worry.

cianfa72
Thus we end up there exist an electric potential difference between the two battery electrodes (it basically equals the battery emf when the external conductive circuit is open) and the "source" for the electric field inside it is due basically to an excess of charges located onto the electrodes themselves.

Come back to the original question and consider for instance the battery negative pole. If there exist an excess of electrons on it (even if, as said before, the overall battery net charge is nevertheless zero) why the meter connected between it and the ground (Earth) will not flow any current through it measuring a non-zero voltage ?

Gold Member
Thus we end up there exist an electric potential difference between the two battery electrodes (it basically equals the battery emf when the external conductive circuit is open) and the "source" for the electric field inside it is due basically to an excess of charges located onto the electrodes themselves.

Come back to the original question and consider for instance the battery negative pole. If there exist an excess of electrons on it (even if, as said before, the overall battery net charge is nevertheless zero) why the meter connected between it and the ground (Earth) will not flow any current through it measuring a non-zero voltage ?
If the meter allows any current to flow at all then that terminal will end up at Earth Potential and the other terminal will be above or below 0, depending on which terminal has been grounded.
But, before any connection has been made, the mean potential of the battery is totally unknown.
Your problem seems to be that you are mixing real and ideal situations in your question (and your thinking?)

cianfa72
If the meter allows any current to flow at all then that terminal will end up at Earth Potential and the other terminal will be above or below 0, depending on which terminal has been grounded.
The meter (voltmeter) has an high impedance nevertheless it allows current to flow. Thus, connecting it for instance between battery negative pole and the ground, I believe the excess electrons on the negative electrode will flow though the voltmeter towards the ground (we assume Earth basically is able to absorb or "pump" electrons without changing its potential). It that is true why it will not measure any voltage ?

But, before any connection has been made, the mean potential of the battery is totally unknown
Why you take in account the "mean" potential of the battery and not the potential of the negative terminal that will be grounded through the voltmeter in the experiment we are considering ?

If there exist an excess of electrons on it (even if, as said before, the overall battery net charge is nevertheless zero) why the meter connected between it and the ground (Earth) will not flow any current through it measuring a non-zero voltage ?

Where does this excess charge come from? If there was an excess charge unless this excess charge is replenished simultaneously as flows to ground through the meter the flow would progress at the speed of light and would have to be sustained which would require what source of electrons? If electrons leave an equal quantity of positive charges would be left to prevent further electrons from leaving.

Gold Member
The meter (voltmeter) has an high impedance nevertheless it allows current to flow. Thus, connecting it for instance between battery negative pole and the ground, I believe the excess electrons on the negative electrode will flow though the voltmeter towards the ground (we assume Earth basically is able to absorb or "pump" electrons without changing its potential). It that is true why it will not measure any voltage ?

No, this is still incorrect.
As has been pointed out several times in this thread, nothing will flow as there is no complete circuit back to the + terminal of the battery

Periwinkle
The meter (voltmeter) has an high impedance nevertheless it allows current to flow. Thus, connecting it for instance between battery negative pole and the ground, I believe the excess electrons on the negative electrode will flow though the voltmeter towards the ground (we assume Earth basically is able to absorb or "pump" electrons without changing its potential). It that is true why it will not measure any voltage ?

Why you take in account the "mean" potential of the battery and not the potential of the negative terminal that will be grounded through the voltmeter in the experiment we are considering ?

I tried it today. I connected a 1.5 V battery with a small bulb and measured with a multimeter. The voltage difference between the two ends of the dry battery was 1.5 V. When the bulb burned, I noticed it was slightly less, about 1.4 V. I tried to find Earth potential - of course I didn't use the ground contact of the electrical socket - but the bathroom faucet, outside the terrace with the iron barrier, the iron column of the clothes dryer. Whether the electric bulb was burning or not, the multimeter did not show any electrical voltage to the ground. Absolute 0.

I've studied Sir Lawrence Bragg's excellent book, Electricity, which I've read before. In my opinion, the multimeter does not show a potential difference compared to the ground - you look at any pole, and whether there is current or not - because it is not. However, not only is there no difference in potential compared to the ground, but it is not between the battery plus and the minus poles. As L. Bragg explains, the electric current flows from the copper electrode to the zinc electrode outside the battery, while in the liquid the positive ions flow from the zinc to the copper electrode. Moreover, Bragg also asks the question: which electrode has greater potential?

In my opinion, this is a chemical process. Copper binds electrons better, so if the two electrodes are connected by a wire, the electrons start to flow into the copper. The zink electrode begins to decompose, and since the electron transfer causes the copper electrode to be negatively charged, zinc and copper ions with a positive charge in the liquid increase the copper electrode. Zinc is slowly running out, and the battery is exhausted. Thus, as a result of the chemical process, the copper electrode absorbs electrons from the zinc, the device shows an electric current, while there is, in fact, no potential difference between the two electrodes. (The migration of positive ions evidences this.)

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thomasj