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

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.
 

BvU

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No, it will measure noise
 

sophiecentaur

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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?)
 
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.
 

BvU

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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.
 
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.
 
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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BvU

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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
 
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..
 

BvU

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No. @Periwinkle has muddled your thread by bringing in the confusing
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) ?
 

gleem

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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.
 
No. @Periwinkle has muddled your thread by bringing in the confusing
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?
 

berkeman

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sophiecentaur

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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.
 
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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sophiecentaur

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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.
 
No. @Periwinkle has muddled your thread by bringing in the confusing
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.
 

davenn

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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
 

davenn

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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
 
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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sophiecentaur

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"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.
 
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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sophiecentaur

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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.
 

gleem

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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.
 
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 ?
 

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