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

sophiecentaur

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instantaneously
You should avoid that word here. If you are trying to really get to the bottom of this then you need to assume that there are time constants involved; everything involves some delay. You have a tiny C (perhaps a few pF) discharging a tiny charge through a very high meter resistance (say 100MΩ) which means a time constant (RC) of less than 1ms but certainly not 0s. The charge flow would be probably enough to pick up as a small click on a radio receiver, places near it.
 
You should avoid that word here. If you are trying to really get to the bottom of this then you need to assume that there are time constants involved; everything involves some delay. You have a tiny C (perhaps a few pF) discharging a tiny charge through a very high meter resistance (say 100MΩ) which means a time constant (RC) of less than 1ms but certainly not 0s. The charge flow would be probably enough to pick up as a small click on a radio receiver, places near it.
You're really talking about the first case (A) I guess. As highlighted several times in this thread, in the second case (B) there will be no current flow (no discharging) at all, nevertheless the voltage difference between plate 2 and Ground will drop to 0 (even with a voltmeter inserted in line) upon turning on the switch (perhaps in some pico-nano seconds)
 

sophiecentaur

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You're really talking about the first case (A)
But the second case is indeterminate - unless there is specific PD quoted, it could be anything. That's been my point all along. It's an equation with two unknown values. The actual PD of that arrangement, floating in space could be anything and you don't know its history.
 
But the second case is indeterminate - unless there is specific PD quoted, it could be anything. That's been my point all along. It's an equation with two unknown values. The actual PD of that arrangement, floating in space could be anything and you don't know its history.
ok but, whatsoever the potential difference of that arrangement is, why a voltmeter connected between the plate and the ground (case B) should not measure anything (not even a small variation/reading provided an enough slow discharge with an adequate time constant ) ?
 
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sophiecentaur

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ok but, whatsoever the potential difference of that arrangement is, why a voltmeter connected between the plate and the ground should not be able to measure anything ?
A good enough voltmeter could measure the PD between any two points, anywhere. But why is that of interest in a very practical context like this one? Why would a PSU manufacturer be wasting time producing a device with 'infinite' insulation and with total EMI shielding?
This thread is about real EE unless, as I pointed out way back, you actually specify all the details of the circuit; in which case there could be an answer available. But you keep shifting the goal posts about what you want to know so there is no full answer available yet.
Try to 'follow the rules' of basic circuit theory and go along with how a Voltage Source behaves. Then, once you are familiar with them and can answer basic questions about simple ideal circuits, try introducing extra Resistances into the (non-ideal) circuit - 'near zero" series R and 'very high' parallel R and see what you get.
 
A good enough voltmeter could measure the PD between any two points, anywhere. But why is that of interest in a very practical context like this one? Why would a PSU manufacturer be wasting time producing a device with 'infinite' insulation and with total EMI shielding?
This thread is about real EE unless, as I pointed out way back, you actually specify all the details of the circuit; in which case there could be an answer available. But you keep shifting the goal posts about what you want to know so there is no full answer available yet.
I would like emphasize that this is a "theoretical question" coming from my doubt about the physics behind the initial scenario shown
 
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davenn

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ok but, whatsoever the potential difference of that arrangement is, why a voltmeter connected between the plate and the ground (case B) should not measure anything (not even a small variation/reading provided an enough slow discharge with an adequate time constant ) ?

As you have been told a number of times
There is an incomplete circuit, therefore NO CURRENT FLOWS

Repeating the same erroneous Q over and over isn't going to get a different result.
 
As you have been told a number of times
There is an incomplete circuit, therefore NO CURRENT FLOWS
I see that, but try to consider the following point of view: if the whatsoever initial potential of the electrode respect to Earth has to drop to the Earth potential (let's assume 0V) upon connecting the voltmeter to the ground, a rearrangement of charges on the electrode has to take place I guess...does not that imply a flow of charges (electrons) somehow ?
 

davenn

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if the whatsoever initial potential of the electrode respect to Earth has to drop to the Earth potential

Why do you think it has to drop ??

I don't see any reason why it should

upon connecting the voltmeter to the ground, a rearrangement of charges on the electrode has to take place I guess
You have to stop guessing. Do the experiment, see what happens


No complete circuit ... No current flows
 

davenn

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upon connecting the voltmeter to the ground, a rearrangement of charges on the electrode has to take place I guess...does not that imply a flow of charges (electrons) somehow ?

That would only happen with a static electric charge ....
eg
You walk across the carpet and you accumulate charge on your body
You touch the metal door knob and get a shock as the accumulated charge returns to where it
came from


Tho you may think it's just a one way circuit because you don't consider the accumulation of charge
as you walk, never-the-less that is part one of the circuit in action
Part two comes in to action when you discharge as you touch the door knob ... the circuit is completed


Even tho it is after sunset here and I know what the result will be, I'm going to do the experiment
using the 12V car battery ... If it was daylight, I would video the results for you.... maybe another day

I'm really starting to wonder if you have this static discharge and charge redistribution stuck
in your mind and you
somehow think it also applies to any situation including a battery ( DC) situation ?
Your continued non-acceptance of the physics, as has been presented, suggests such ??


Dave
 

gleem

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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 earths 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.
 
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 earths 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...
 
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sophiecentaur

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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

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. 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
 
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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

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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.
 
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:
Pres1.png

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

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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

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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.
 
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:

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

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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:

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

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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.
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

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

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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.
 
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