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

[davenn]

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,

ohhh and I forgot to mention earlier...
A voltmeter doesn't go in series with a circuit

 

[davenn]

Wise words...
View attachment 242150
This is the basic schema we are discussing about using for instance a simple Daniell cell. Up to now we have defined no electrical model for it...can you propose one ?

Ok, that diagram just confirms previous comments by others and myself, there will be no current flow to GND

 

[davenn]

Wise words...
View attachment 242150
This is the basic schema we are discussing about using for instance a simple Daniell cell. Up to now we have defined no electrical model for it...can you propose one ?

Now change it to this configuration and it's a whole different story

Dave

 

[cianfa72]

Ok, that diagram just confirms previous comments by others and myself, there will be no current flow to GND

ok, but...consider an initially isolated battery (e.g. Daniell cell): the zinc electrode has got an accumulated amount of (excess) electrons whereas the copper one an excess of positive charges (it turns out that when a sufficient amount of negative (positive) charges have been accumulated on the electrodes the chemical process will stop). There exist, thus, a voltage difference between them and (I suppose) between each of them and the ground (Earth).

Next, suppose at a given point in time you connect the zinc electrode (-) to the ground (Earth) through a voltmeter: could you explain which is the reason why no electrons will flow from the zinc electrode towards the Earth ?

 

[gleem]

I think there is a problem of labeling the electrodes with a + or - sign for it might imply to some that the electrode are a source or a bank of charges. It might be better to refer to them as the anode or cathode instead. Just a thought.

 

[gleem]

The excess of charge before the electrode are connected is due to the inter electrode capacitance which is of the order of picofarads. This might accumulate about 10⁷ electrons. The potential difference of the cell is not due to the excess charges on the electrode for if it were then the physical size and placement of the electrodes in the cell would change the voltage across the cell which does not happen. If you ground the cathode no electrons will flow off it to ground because they are being held in place by the induced positive charges on the anode. Grounding the cathode does not change the potential difference between the electrodes.

 

[cianfa72]

The excess of charge before the electrode are connected is due to the inter electrode capacitance which is of the order of picofarads. This might accumulate about 10⁷ electrons.

ok, good point

The potential difference of the cell is not due to the excess charges on the electrode for if it were then the physical size and placement of the electrodes in the cell would change the voltage across the cell which does not happen.

Not sure to understand your point: for sure the electromotive force (emf) of a cell is determined by characteristics of chemical processes involved and, from the very definition of emf, it equals the potential voltage difference established across electrodes when the cell is "open" (no external conductive path between electrodes)

If you ground the cathode no electrons will flow off it to ground because they are being held in place by the induced positive charges on the anode.

Could you please elaborate a bit this point ? Thanks

 

[rcgldr]

Although the amount of charge at the terminals is small, it seems that you could have a wire connected to an isolated plate or sphere that retains charge, and alternately connect the wire between the two terminals of an isolated battery (connecting only to one terminal at a time), to very slowly discharge a battery. I don't know if this could be considered as some type of intermittent circuit.

 

[gleem]

Not sure to understand your point: for sure the electromotive force (emf) of a cell is determined by characteristics of chemical processes involved and, from the very definition of emf, it equals the potential voltage difference established across electrodes when the cell is "open" (no external conductive path between electrodes)

When you place a capacitor across the terminals it will charge to the potential of the battery. Only enough charge will be created by the chemical process to charge the battery to its emf according to this relationship'

Q = CV_emf

If I connect a 1 uF cap and it charges to Q if I change the cap to 10uF the charge now will be 10Q because the voltage is the same.

The two electrodes form a capacitor so if the electrode size, relative distance from one another is changed the capacitance C will change resulting in different amount of charge.

Could you please elaborate a bit this point ? Thanks

Not much more. Giving an alternative path for the electrons like connecting the cathode to ground does nothing because the positive charges on the other electrode are attracting the electrons more than the ground so they stay put.

 

[davenn]

ok, but...consider an initially isolated battery (e.g. Daniell cell): the zinc electrode has got an accumulated amount of (excess) electrons whereas the copper one an excess of positive charges (it turns out that when a sufficient amount of negative (positive) charges have been accumulated on the electrodes the chemical process will stop). There exist, thus, a voltage difference between them and (I suppose) between each of them and the ground (Earth).

it's irrelevant ... for every - charge on the - electrode, there is a positive charge on the + terminal ... there is a balance
and an electric field that exists in the cell/battery between the 2 terminals
Adding a ground to either terminal IS NOT going to make any difference.

Next, suppose at a given point in time you connect the zinc electrode (-) to the ground (Earth) through a voltmeter: could you explain which is the reason why no electrons will flow from the zinc electrode towards the Earth ?

see my previous comment

 

[cianfa72]

it's irrelevant ... for every - charge on the - electrode, there is a positive charge on the + terminal ... there is a balance and an electric field that exists in the cell/battery between the 2 terminals
Adding a ground to either terminal IS NOT going to make any difference.

ok thanks, I believe got it.

Thus, just to check my understanding, the following two cases are actually different.

In the first case (A) at the given moment conductor 2 (basically a plate of a capacitor) is grounded positive charges discharge towards Earth (or in equivalent way electrons from the Earth neutralize them) whereas in the second case (B) electrons on plate 2 do not discharge because of the "attraction/electrostatic induction" due to positive charges existing on plate 1. In any case electric potential of plate 2 reaches the same as Earth.

Does it make sense ?

 

[sophiecentaur]

@cianfa72
Your last paragraph could benefit from a switch to be included in the circuit, to make clear what is happening. IAnd, if you want to be really rigorous about your description, you could probably include a first diagram in the series, when t = 0 and the switch is open.

 

[davenn]

In the first case (A) at the given moment conductor 2 (basically a plate of a capacitor) is grounded positive charges discharge towards Earth (or in equivalent way electrons from the Earth neutralize them)

You are not going to get case A for the very reasons I stated in my previous post

you are not going to get + and - charges on the same terminal like that ...
the charge there is going to be equal and opposite to what ever is on the other terminal ... period

 

[sophiecentaur]

you are not going to get + and - charges on the same terminal like that ...

There is nothing wrong with drawing + and - charges on one plate. The resulting NET charge is what will count in the end. (It could perhaps have been better to draw two columns of + charges on the + plate to make this clear.

 

[cianfa72]

There is nothing wrong with drawing + and - charges on one plate. The resulting NET charge is what will count in the end. (It could perhaps have been better to draw two columns of + charges on the + plate to make this clear.

ok, as advised I re-drew the pictures as follows:

Case A is just to help reasoning and does not represent the original scenario.

We can conclude that, upon turning on the switch, instantaneously the voltage difference between the plate and Ground will drop to 0 (as in the case there was a voltmeter inserted in series with the switch: upon switching it on instantaneously the initial difference voltage would drop to 0 too)

 

[sophiecentaur]

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.

 

[cianfa72]

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]

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.

 

[cianfa72]

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 ) ?

 

[sophiecentaur]

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.

 

[cianfa72]

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

 

[davenn]

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.

 

[cianfa72]

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]

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 happensNo complete circuit ... No current flows

 

[davenn]

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 into action when you discharge as you touch the door knob ... the circuit is completedEven 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]

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.