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

[gleem]

Let me try this. When you connect the cathode to Earth it is like connecting another capacitor between the electrodes. The Earth is one plate and the anode the other. When you do this you are just making the cathode much larger. But the charge that this new capacitor can accommodate is still small because the anode is still small and the overall capacitance is only about twice the inter electrode capacitance. For a 9 V battery I estimated it is of the order of about 10^-14F. Connecting a voltmeter with 100 Mohm input impedance gives a charging time constant for this setup of about 1usec. When charged the Earth has acquired about 500,000 electrons (produced by the chemical reaction in the battery). A typical 9 volt battery has a capacity to generate about 300 mAH or 62x10²¹electrons. So yes a current flowed but lasted too short a time for a typical multimeter to register. However if you put an oscilloscope from the cathode to Earth with a switch in series and set your time base to say 0.5 usec/cm when the switch is closed I would expect to see an exponentially decreasing signal going to zero by the end of the trace. So for all intents and purposes connecting an electrode to ground has little effect and certainly cannot sustain a current. The Earth's potential is irrelevant as long as there are not conductive paths to the anode.

If you take you battery to the top of a power line and carefully connect the cathode to it nothing happens just like the bird who is sitting next to it.

 

[cianfa72]

Let me try this. When you connect the cathode to Earth it is like connecting another capacitor between the electrodes. The Earth is one plate and the anode the other. When you do this you are just making the cathode much larger. But the charge that this new capacitor can accommodate is still small because the anode is still small and the overall capacitance is only about twice the inter electrode capacitance. For a 9 V battery I estimated it is of the order of about 10^-14F. Connecting a voltmeter with 100 Mohm input impedance gives a charging time constant for this setup of about 1usec.

ok, basically you are considering the series of voltmeter input impedance and capacitor to work out the charging time constant

So yes a current flowed but lasted too short a time for a typical multimeter to register. However if you put an oscilloscope from the cathode to Earth with a switch in series and set your time base to say 0.5 usec/cm when the switch is closed I would expect to see an exponentially decreasing signal going to zero by the end of the trace. So for all intents and purposes connecting an electrode to ground has little effect and certainly cannot sustain a current. The Earth's potential is irrelevant as long as there are not conductive paths to the anode.

Sure, make sense: I'm not able to check with an oscilloscope but it would be very interesting...

 

[sophiecentaur]

I'm not able to check with an oscilloscope

It's one of those things you don't really need to check. If you assume things like the Capacitance and Resistance of the scope probe and the likely capacitance of the battery and the 9V - or whatever, you can predict quite accurately enough the result.

that is part one of the circuit in action

There's a little too much emphasis on the importance and relevance of the 'circuit' word. Whilst an object acquires a charge there is an implied part of a circuit and whilst it discharges, there is the remainder of the circuit. It's not a 'DC' circuit problem so the 'circuit' needn't be complete all at the same time.

 

[davenn]

. It's not a 'DC' circuit problem so the 'circuit' needn't be complete all at the same time.

The OP's problem is a DC circuit problem and therefore needs a complete circuit and he seems to be having
great problems understanding that point.
And the point I was making with the static example was, if you really did read it, that he seems to be confusing the 2 different situations
by trying to apply a static situation to a DC situation

 

[cianfa72]

The OP's problem is a DC circuit problem and therefore needs a complete circuit and he seems to be having great problems understanding that point.

I do not think that: to me a DC (Direct Current) problem occurs when no dynamic evolution take place (included in DC problems are the stationary current cases).
In our case, upon turning on the switch, there could be variations of the electrical quantities involved too

 

[sophiecentaur]

And the point I was making with the static example was, if you really did read it, that he seems to be confusing the 2 different situations

I agree. It basically is not a DC scenario, despite how the thread started off and there is no answer to the question without a full description of the initial conditions. This is daft because we 'all' (at least both) understand the situation and we are just trying to explain away an unreal situation. I really wish that people (eg the OP) would go to these problems via a proper learning of basic EM theory. The basics always deliver the right answers when they are used properly and you can't leap in half way.

 

[cianfa72]

I agree. It basically is not a DC scenario, despite how the thread started off and there is no answer to the question without a full description of the initial conditions. This is daft because we 'all' (at least both) understand the situation and we are just trying to explain away an unreal situation.

I apologize if I am not able to explain the point or my English is not so good. About the initial conditions: to me they look clear as depicted again in the following picture:

If you prefer this time the instrument is an oscilloscope having let's say 100 Mohm input impedance and the battery is just a Daniell cell (emf 1.10V). Which other initial conditions are needed to describe what happens upon turning on the switch ? I believe the point here is that, up to now, we have not a definitive model for it (for instance which is the model to include the capacitance involved ?)

 

[sophiecentaur]

I apologize if I am not able to explain the point or my English is not so good. About the initial conditions: to me they look clear as depicted again in the following picture:
View attachment 242452
If you prefer this time the instrument is an oscilloscope having let's say 100 Mohm input impedance and the battery is just a Daniell cell (emf 1.10V). Which other initial conditions are needed to describe what happens upon turning on the switch ? I believe the point here is that, up to now, we have not a definitive model for it (for instance which is the model to include the capacitance involved ?)

You just made my point here. If you ‘turn on a switch’ you do not have a DC situation. You do not need to specify the measuring instrument. They all have a similar input stage, in any case, these days.
The potential is whatever it started off as (arbitrary). If you discharge by connecting via any conductance. The potential of the connected terminal will be 0V eventually.

 

[sophiecentaur]

the battery is just a Daniell cell (emf 1.10V).

Why are you getting so specific about the equipment and battery and why use such a low voltage ( and possibly unavailable) battery. An AA battery would be just as useful.
I am still not sure if you understand that all you are likely to see will depend more on the arbitrary volts that the isolated battery can have acquired as static charge than any PD across the terminals. You seem to want to 'prove something' but any blip you get on the scope will not really prove anything. Also, there could well be some induced AC volts (probably due to the 'scope leads as much as anything). Frankly, it sounds a bit of a nonsense experiment. Experiments with no preparation or prediction, based on some theory tend to be a bit of a waste of time, however entertaining they may be.

 

[mselectromagnetic]

Since the battery has both inductance and capacitance, maybe this article by Tesla will help answer the question of whether a potential exists in relation to Earth ground:
https://www.thomastownsendbrown.com/petro/teslacap.htm
If there was anybody who understood this kind of thing, it was Tesla. What I think would be interesting would be to measure the potential of the battery's NEGATIVE pole in relation to Earth ground...at different elevations above the earth. In solar energy systems, the common practice is to use Earth ground to be compliant with code in most parts of the country. However, I bet you would never see the negative pole of the battery bank grounded in similar fashion...because there might actually be a parasitic potential there.

 

[sophiecentaur]

Since the battery has both inductance and capacitance, maybe this article by Tesla will help answer the question of whether a potential exists in relation to Earth ground:
https://www.thomastownsendbrown.com/petro/teslacap.htm
If there was anybody who understood this kind of thing, it was Tesla. What I think would be interesting would be to measure the potential of the battery's NEGATIVE pole in relation to Earth ground...at different elevations above the earth. In solar energy systems, the common practice is to use Earth ground to be compliant with code in most parts of the country. However, I bet you would never see the negative pole of the battery bank grounded in similar fashion...because there might actually be a parasitic potential there.

That's a bit typical of Tesla publications. No actual figures quoted and loads of Astrological style.

 

[cmb]

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.

I like the way the OP asked this question, because they have embedded a little piece of 'natural philosophy' into it.

I will rephrase it to pose the question in a slightly different way; All isolated objects have their own distinct electric charge, which is (at least, theoretically) measurable compared to any other distinct object. So is the difference between the electric charges between two objects not a measure of 'potential difference'? If so, we know one terminal on the battery has more electrons than the other end, so we also know that there is more charge on one of the battery terminals to the other. So, "whatever that difference is" to ground, even if it is some imaginary number, the 'other terminal' should be a larger/smaller difference by 9V?

I will answer this in the following way;-
"Potential difference" is measured in Volts. Volts are Joules per coulomb of charge.

You are correct that the difference (to Earth) in the amount of charge on one battery terminal will be greater than the other terminal, whether or not that is practically measurable is a different question. Therefore, logically, it is inescapable that you would also be correct that one terminal has a bigger difference of charge to the other, with respect to any other object.

But here's the crunch - 'potential' means the potential to do work. There is a charge difference, but it is not (necessarily) a difference that can do work. Ergo, not related to 'potential difference'.

I will add one caveat on that, because it is not precisely true that 'zero' work is done, it is simply virtually unmeasurable in the scenario you mention. What happens when you connect your voltmeter to the negative terminals and ground is that you push up the positive terminal to +9V above ground by doing that. Likewise if you did that to the +ve you'd push the negative terminal low. In doing so, you would change the energy stored in the 'battery-Earth' capacitor system, because there would be some non-zero finite capacitance, thus energy stored. As the battery pushes its terminals up and down voltage relative to ground, it would 'do work'. But as the capacitance would be so small, any physical instrument would swamp that signal.

If, however, you connected one of the terminals to an actual capacitor, and the other end of the capacitor to ground, it would still not be connected to ground. But if you put your voltmeter on one of the terminals now, you'd likely see the meter giving an actual real reading, and it would likely change over time tending to zero as the capacitor charges up to the battery voltage. In practice with no capacitor, this happens virtually instantaneously because the battery-Earth capacitance is so small. The time taken will be according to the 'RC' curve where R is your voltmeter internal resistance.

So I would refute that the potential difference on an isolated battery terminal is undefined with respect to any other object. There has to be absolutely zero capacitance between the two objects for there to be no potential difference. OK, so that might be 10^-100 small and overwhelmed by even cosmic radiation coming along and upsetting the charges on the parts, but there will be a non-zero potential so long as there is a capacitance between them.

It will, of course, be completely unmeasurable by the OPs voltmeter, so that answers the question!

 

[BvU]

Let's ask @cianfa72 if that answers the question ?

 

[cianfa72]

What happens when you connect your voltmeter to the negative terminals and ground is that you push up the positive terminal to +9V above ground by doing that. Likewise if you did that to the +ve you'd push the negative terminal low. In doing so, you would change the energy stored in the 'battery-Earth' capacitor system, because there would be some non-zero finite capacitance, thus energy stored. As the battery pushes its terminals up and down voltage relative to ground, it would 'do work'. But as the capacitance would be so small, any physical instrument would swamp that signal.

in this model which are actually the two 'plates' of the 'battery-Earth' capacitor system ?

If, however, you connected one of the terminals to an actual capacitor, and the other end of the capacitor to ground, it would still not be connected to ground. But if you put your voltmeter on one of the terminals now, you'd likely see the meter giving an actual real reading, and it would likely change over time tending to zero as the capacitor charges up to the battery voltage. In practice with no capacitor, this happens virtually instantaneously because the battery-Earth capacitance is so small. The time taken will be according to the 'RC' curve where R is your voltmeter internal resistance.

Not sure to understand the model: how is the voltmeter connected here ? Are you considering a simple 'RC' series circuit ?

Thanks

 

[Nugatory]

in this model which are actually the two 'plates' of the 'battery-Earth' capacitor system ?

One of them is the battery and the other is the earth. (Don't be confused because neither of them look like "plates". Introductory textbooks generally work with the special case of parallel plate capacitors but that's not because capacitors have to be that way, it's because the parallel plate capacitor is an easy example of how to calculate approximate capacitance from first principles).

Not sure to understand the model: how is the voltmeter connected here ? Are you considering a simple 'RC' series circuit ?

No, we are not considering a simple RC circuit. After three pages of this thread, do we have say, yet again, that THERE IS NO CIRCUIT?

We have a capacitor with one terminal connected to the battery and the other terminal connected to the earth, and a voltmeter with its probes on each side of the capacitor. And this is equivalent to the configuration you started with, one probe on the battery and the other on the earth; it's only interesting to consider it because it's easy to buy a capacitor with much more capacitance than in your original setup.

 

[cmb]

in this model which are actually the two 'plates' of the 'battery-Earth' capacitor system ?

The metallic conductor of each electrode in the battery.

In practice (if 'in practice' makes any sense in this scenario with virtually zero actual capacitance), assuming the battery case is metallic which it probably is, there will be a capacitance between the outer case of the battery and Earth (the two 'plates') then you will effectively have two further capacitors in parallel (but in series with the case-Earth 'capacitance') to the battery electrodes

Not sure to understand the model: how is the voltmeter connected here ? Are you considering a simple 'RC' series circuit ?

Thanks

An ideal voltmeter has infinite resistance. But it is impossible for a real world voltmeter to have infinite resistance. They usually come with 1 MOhm or 10MOhm internal resistance, so that'll be your 'R'.

[In fact, I often measure kV/ua stuff with cheap multimeters (they tend to blow up on a regular basis, no point buying expensive ones) and I don't bother with the current selections, I just pass the current straight through the 'voltmeter' setting, giving me ua = volts based on a 1 MOhm internal resistance.]

FWIW There is no such thing as an infinite resistance. It's funny really, because intuitively you'd expect 'something' to be a perfect insulator and 'nothing' to be a perfect conductor, but reality treats us to something non-intuitive that there is no such thing as a perfect insulator, but there is such a thing as a perfect conductor! Your voltmeter will always pull the battery terminal to ground when you try to measure it, there is nothing connected to the other terminal to stop that. A capacitance would slow that down, but not stop it.

 

[cianfa72]

No, we are not considering a simple RC circuit. After three pages of this thread, do we have say, yet again, that THERE IS NO CIRCUIT?

yes got it...there is no galvanic closed circuit through the Earth.

We have a capacitor with one terminal connected to the battery and the other terminal connected to the earth, and a voltmeter with its probes on each side of the capacitor. And this is equivalent to the configuration you started with, one probe on the battery and the other on the earth; it's only interesting to consider it because it's easy to buy a capacitor with much more capacitance than in your original setup.

here, let me say, we are basically considering a simple RC parallel where R is actually the internal voltmeter resistance

 

[mselectromagnetic]

That's a bit typical of Tesla publications. No actual figures quoted and loads of Astrological style.

I guess he was writing for the general public at the time. However, I have seen Youtube videos of people putting wires at the top of a pine tree and measuring a potential when putting a meter from that lead to a grounded lead. I don't recall seeing any astrology in that article.

 

[gleem]

I have seen Youtube videos of people putting wires at the top of a pine tree and measuring a potential when putting a meter from that lead to a grounded lead.

Could you find it URL so we could see it?

 

[mselectromagnetic]

Could you find it URL so we could see it?

 

[mselectromagnetic]

This is not the one I saw before, but still somewhat interesting. The other one will take some searching, but I think it was John Hutchison doing the experiment.

 

[sophiecentaur]

I don't recall seeing any astrology in that article.

I referred to astrological style and not astrology. Tesla did an awful lot of waffling and had his audience enthralled. I'm amazed that the 'enthraldom' seems to persist to this day. True, he made a choice about adopting the use of AC Power; lucky guess? There was a lot of 'general ignorance' about 'electricity' at the time and fortunes were won and lost on the strength of non-engineering hunches. ("So what's different now?", I hear you cry - but PF didn't exist in those days)
We haven't been shown that article but another article . . . . . .

This is not the one I saw before, but still somewhat interesting. The other one will take some searching, but I think it was John Hutchison doing the experiment.

He says he's measuring "nine point nine volts" when the meter appears to read 0.99V. He has actually posted it! What does that say about experimental method? How long did the PD exist for and what was the drain current?

 

[hutchphd]

He says he's measuring "nine point nine volts" when the meter appears to read 0.99V. He has actually posted it! What does that say about experimental method? How long did the PD exist for and what was the drain current?

Looks like he is using galvanized screws. I believe the emf of copper- zinc galvanic cell is about 1V. The man is an idiot.

 

[Jyothish]

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.

Answer to your question in one word is "no"
To clarify further we may split the question into two parts
1) Do you think there will exist a electric potential difference(between +clamp and earth)? Answer is "yes"
2) Can the above potential difference be measured by voltmeter? Answer is "no".Because voltmeter needs a closed circuit to work