I Potential difference between a battery's terminal and Earth ground

[cianfa72]

TL;DR Summary

About the definition of electric potential difference between a battery's terminal and a connection to Earth's ground.

We had a similar thread some time ago. I'd like to reconsider some aspects.

Consider a 9V battery just to fix ideas. The chemical electromotive force inside it establishes an amount of positive charges on its positive terminal and an exact same amount of negative charges on its negative terminal.

Suppose the battery is at 1mt from the Earth surface. The charge distribution on the Earth plus the charge distribution on the battery terminals will be the sources of a static conservative electric field $E$. Hence one can calculate the elettric potential evaluating its difference for instance between the battery's positive terminal and the Earth. So far so good.

Now suppose to attach a voltmeter between the battery's positive terminal and a Grounded metal rod. The voltmeter will not measure any voltage since it requires a current flow even though its ingress impedence in very high.

In any case I believe there is a voltage difference between them even though a voltmeter cannot measure it.

Does it make sense? Thanks.

 

[Baluncore]

In any case I believe there is a voltage difference between them even though a voltmeter cannot measure it.

Does it make sense? Thanks.

No.

If the voltmeter has a real internal resistance, measuring the voltage between ground and the floating battery will rapidly bring the voltage of the battery terminal down to ground, with a reading of zero volts.

If the voltmeter has infinite resistance and zero current, no current will flow through the voltmeter, which will read whatever floating voltage was on the battery terminal, relative to ground, when the meter was connected.

 

[cianfa72]

If the voltmeter has a real internal resistance, measuring the voltage between ground and the floating battery will rapidly bring the voltage of the battery terminal down to ground, with a reading of zero volts.

Ok, since there is no electromotive force driving a current through the voltmeter, it will read zero volt.

If the voltmeter has infinite resistance and zero current, no current will flow through the voltmeter, which will read whatever floating voltage was on the battery terminal, relative to ground, when the meter was connected.

Yes, however an actual voltmeter will always have a finite internal resistance, I believe.
Does exist a voltmeter with infinite internal resistance/impedance and zero current?

 

[Baluncore]

Yes, however an actual voltmeter will always have a finite internal resistance, I believe.
Does exist a voltmeter with infinite internal resistance/impedance and zero current?

Yes. A field mill.
https://en.wikipedia.org/wiki/Field_mill

 

[cianfa72]

Yes. A field mill.
https://en.wikipedia.org/wiki/Field_mill

Ah ok, it basically "executes" the line integral of the electric field between two points evaluating the potential difference.

 

[cianfa72]

I'm confused about the following: when the voltmeter is connected accross the battery's terminal and ground will there be a temporary flow of charges/electrons through it ?

 

[Baluncore]

I'm confused about the following: when the voltmeter is connected accross the battery's terminal and ground will there be a temporary flow of charges/electrons through it ?

It all depends on what type of voltmeter you are referring to.
Probably, yes.
The temporary return circuit would be through capacitance between the terminal and ground. You are discussing silly incomplete circuits without return conductors, measuring voltages between floating things, that should be grounded to something sensible.

 

[cianfa72]

It all depends on what type of voltmeter you are referring to.
Probably, yes.
The temporary return circuit would be through capacitance between the terminal and ground.

I refer to a "standard" voltmeter with a finite ingress impedance. I missed the point: if there was a flow of electrons inside it, would the charges on the battery's terminal actually be replenished by the battery's electromotive force?

 

[Baluncore]

I refer to a "standard" voltmeter with a finite ingress impedance.

A voltmeter is used to measure the voltage difference between two conductors that are part of a complete circuit.
Your components are floating. You have no circuit.

 

[Baluncore]

I missed the point: if there was a flow of electrons inside it, would the charges on the battery's terminal actually be replenished by the battery's electromotive force?

What is it?
What is the circuit?

 

[cianfa72]

This is the schematic: there is no closed circuit at all.

My point is simple: even though there is no external circuit connecting the battery's terminals, it still establishes an amount of negative charges at $-$ terminal and an equal amount of positive charges at the $+$ terminal. What happen when the switch is suddenly turned on ?

We said that there is a potential difference between Ground and the $-$ terminal due to a static electric field produced by "sources" consisting of charges located on the battery's terminals and on the Ground. Therefore I believe there will be a temporary flow of electrons upon the switch is turned on.

Is that is the case, will the chemical processes inside the battery replenish the "lost" electrons at the $-$ terminal ?

 

[Dale]

it still establishes an amount of negative charges at − terminal and an equal amount of positive charges at the + terminal

This is not necessarily true. It establishes a potential difference between the terminals, but that needn’t imply equal (and opposite) charges on the terminal.

This is the schematic: there is no closed circuit at all.

Any “floating” conductor can be approximated by a small capacitative connection to ground. Then you have a valid circuit you can analyze

 

[Baluncore]

Is that is the case, will the chemical processes inside the battery replenish the "lost" electrons at the terminal ?

The battery is a fixed voltage source. You cannot get charge out of one end, without putting charge in at the other end.

What is the voltage on the negative terminal of the flying battery, relative to ground, before the ground switch is closed?

 

[Baluncore]

Without battery terminal capacitance to ground, there will be no battery current flowing during the flying to grounded transition.

If the battery terminals have 2 pF capacitance to ground each, then as the negative terminal approaches ground, so will the positive terminal. The 2 pF on the positive terminal will be discharged through the battery, the same current will flow from the negative terminal, along with the negative terminal capacitance current. Those currents will flow through the voltmeter resistance, assumed to be about 10 megohm.

If the positive terminal of the battery starts at +24V, with the negative terminal at +12V, then the current flowing through the battery and meter will be about 0.5 uA. That current will flow for a period of about 2pF * 10Meg = 20 us.

That current has nothing to do with the battery charge, it has everything to do with the terminal capacitance. Replace the battery with a copper wire and exactly the same thing will happen.

 

[cianfa72]

Any “floating” conductor can be approximated by a small capacitative connection to ground. Then you have a valid circuit you can analyze

You mean in the picture the conductor attached to the battery's negative terminal (when the switch is off) has a (small) capacitance w.r.t. ground.

 

[cianfa72]

If the battery terminals have 2 pF capacitance to ground each, then as the negative terminal approaches ground, so will the positive terminal. The 2 pF on the positive terminal will be discharged through the battery, the same current will flow from the negative terminal, along with the negative terminal capacitance current. Those currents will flow through the voltmeter resistance, assumed to be about 10 megohm.

If the positive terminal of the battery starts at +24V, with the negative terminal at +12V, then the current flowing through the battery and meter will be about 0.5 uA. That current will flow for a period of about 2pF * 10Meg = 20 us.

Here is my understanding of what you said

Let's say the battery is 9V and suppose that the positive terminal starts at +24V w.r.t. ground and the negative terminal at +12V w.r.t. ground (is it consistent with a 9V emf battery ?). Each terminal has a 2 pF capacitance w.r.t. ground/earth (so one plate of each capacitor actually represents the respective battery's terminal while the other is the common ground). I placed the resistor R1 in the model to account for battery internal connection.

If the above model makes sense, when the switch is suddenly turned on, there will be a flow of electrons from the terminal plates of capacitor C1 e C2 through the voltmeter towards the ground. Why did you estimate a current flow of 0.5 uA through the battery and the voltmeter ?

That current has nothing to do with the battery charge, it has everything to do with the terminal capacitance. Replace the battery with a copper wire and exactly the same thing will happen.

You mean if one replaced the battery with a copper wire (with a given resistance) as connection between the two terminals (which still have a 2 pF capacitance w.r.t. ground) one would get the same result.

 

[cianfa72]

The battery is a fixed voltage source. You cannot get charge out of one end, without putting charge in at the other end.

Ah ok ! In other words there is no way to "extract" charges from one battery's terminal end without putting back charges into the other end terminal. To do that there must be an external circuit connecting them.

 

[Baluncore]

Why did you estimate a current flow of 0.5 uA through the battery and the voltmeter ?

The RC current decays exponentially. The negative battery terminal capacitor starts out at +12V, so the maximum initial current through the 10meg meter will be 12 / 1e7 = 1.2 uA. The positive terminal capacitance must be discharged through the battery to the voltmeter. That will charge the battery.

You mean if one replaced the battery with a copper wire ... one would get the same result.

Is that not what I wrote?

 

[cianfa72]

The RC current decays exponentially. The negative battery terminal capacitor starts out at +12V, so the maximum initial current through the 10meg meter will be 12 / 1e7 = 1.2 uA. The positive terminal capacitance must be discharged through the battery to the voltmeter.

Ok, therefore though the 10meg voltmeter will flow the sum of two currents: the discharge current from the negative battery terminal "capacitor plate" + the discharge current from the positive battery terminal "capacitor plate". The latter flowing through the battery will also charge it.

Just to check my understanding: in your example of battery negative and positive terminal "capacitor plates" starting at +12V and +24V w.r.t. ground respectively, does that imply that the battery emf was exactly (24-12) = 12V ?

In any case the battery emf is not involved at all (i.e. the charges on the battery terminal "capacitor plates" discharged through this process are not "accumulated" there by battery's emf).

 

[Baluncore]

In any case the battery emf is not involved at all (i.e. the charges on the battery terminal "capacitor plates" discharged through this process are not "accumulated" there by battery's emf).

A battery is NOT a capacitor.
An ideal battery maintains a chemically defined voltage while it is being charged or discharged. The voltage may not change, but the charge is being stored in the chemical cells of the battery. In the example case, discharging the terminal capacitance through the battery, will charge the battery. An ideal battery is a fixed voltage source, with zero series resistance.

A capacitor, on the other hand, has a voltage directly proportional to charge. C = Q / v.
You cannot model a battery as a capacitor.

 

[cianfa72]

The voltage may not change, but the charge is being stored in the chemical cells of the battery. In the example case, discharging the terminal capacitance through the battery, will charge the battery. An ideal battery is a fixed voltage source, with zero series resistance.

Ok, so in post#16 schematic, the resistance R1 must be replaced by an ideal voltage source of 12V emf.

From the differential equation system for this circuit, starting at time t=0 with those voltage levels w.r.t. ground, one can calculate the current through the 10meg meter. As you pointed out, its initial value will be actually 12/1e7 = 1.2 uA (at the start no current from the battery positive terminal "capacitor plate" will flow through the 10meg meter).

Does it make sense ?

 

[Baluncore]

Here is my schematic. Load is drawn on the right.
The initial condition is that V( Vp ) = +24 V; Then V( Vn ) = +12 V.

When the switch is closed at t = 0; Currents flow clockwise.
Note that I( C1 ) = I( C2 ). Capacitors are both discharged through R1.

Initially; I( R1 ) = +12V / 10Meg = 1.2 uA .
Since dv/dt of capacitors are the same; I( R1 ) = I( C1 ) + I( C2 ).
Initially; I( C1 ) = I( C2 ) = I( R1 ) / 2 = 0.6 uA .
Series connection implies; I( V1 ) = I( C1 ).

 

[cianfa72]

The initial condition is that V( Vp ) = +24 V; Then V( Vn ) = +12 V.

When the switch is closed at t = 0; Currents flow clockwise.
Note that I( C1 ) = I( C2 ). Capacitors are both discharged through R1.

Initially; I( R1 ) = +12V / 10Meg = 1.2 uA .
Since dv/dt of capacitors are the same; I( R1 ) = I( C1 ) + I( C2 ).
Initially; I( C1 ) = I( C2 ) = I( R1 ) / 2 = 0.6 uA .
Series connection implies; I( V1 ) = I( C1 ).

Ah ok, you did a simulation using LTspice. C1 and C2 are both discharged through R1 since the current through R1 is always the sum of I(C1) and I(C2).

However my doubt is: who is responsible for the initial "charge" on the battery's positive and negative terminal "capacitor plates" ? I believe it is the battery's emf that build up an excess of electrons at the negative terminal's "capacitor plate" vs a defect of them at positive terminal's "capacitor plate".

 

[Baluncore]

However my doubt is: who is responsible for the initial "charge" on the battery's terminal "capacitor plates" at positive and negative terminal ?

You are required to specify the initial conditions for the simulation. It will be some initial voltage, and everything will follow in an orderly and predictable way from that. How you held the battery, and what shoes you were wearing when you assembled the circuit, is quite irrelevant to the analysis of the discharge process.

Static discharge can damage insulation, so components are grounded through high-value resistors, before and during the assembly process. If there is no DC connection between different parts of the circuit, SPICE will warn you.

We could never allow a flying battery to arrive for assembly with an unspecified electrostatic potential on the terminals.

Rather than electronics, maybe you should be studying law.

 

[cianfa72]

You are required to specify the initial conditions for the simulation. It will be some initial voltage, and everything will follow in an orderly and predictable way from that. How you held the battery, and what shoes you were wearing when you assembled the circuit, is quite irrelevant to the analysis of the discharge process.

Static discharge can damage insulation, so components are grounded through high-value resistors, before and during the assembly process. If there is no DC connection between different parts of the circuit, SPICE will warn you.

I believe that's the point. The SPICE directive .IC V(Vp) = 24V sets the initial potential of node Vp (i.e. of the C1 plate) to 24V w.r.t. ground. Is the associated charge on the C1 plate actually due only to possibly electrostatic process/sources (i.e. it is not the charge stored there from the battery) ?

 

[Dale]

who is responsible for the initial "charge" on the battery's positive and negative terminal "capacitor plates" ? I believe it is the battery's emf that build up an excess of electrons at the negative terminal's "capacitor plate" vs a defect of them at positive terminal's "capacitor plate".

That depends on the previous history, not just the battery EMF. Previous manipulations could have left some overall charge on the battery.

 

[Baluncore]

Is the associated charge on the C1 plate actually due only to possibly electrostatic process/sources (i.e. it is not the charge stored there from the battery) ?

A battery stores charges internally. The small terminal capacitance of any component is determined by the geometry of the terminals, and the placement of those terminals in space. The difference in terminal voltage is an internal property of the voltage source component. The common charge on the whole flying battery component is an external property, quite independent of the internal charge.

The conductive wire in a SPICE simulation, that connects common component terminals, is a single equipotential node of zero size. A node has no series resistance, nor capacitance to ground, unless you choose to include those in your model as discrete components. Currents are summed into a node as a single equipotential point.

The wire, as a node in SPICE, does not have inductive or transmission line properties, nor any current direction, unless you need to subdivide one node into several nodes, and include those subcomponents in your circuit model.

You wanted to model charge on the battery terminals, but that defeats the terminals being zero-size nodes, so we added the terminal capacitance to ground, just to keep you happy. Now you seem to be complaining that the terminal capacitance of a flying battery, has variable geometry, as it flies through undefined electrostatic space, to arrive at the circuit assembly point. SPICE does not model flying components during the assembly process.

SPICE was designed to simulate a specified circuit. At some point, you must stop adding unnecessary complexity to your model, and you must settle for the approximation of the assembled circuit schematic.

 

[cianfa72]

Finally I think I managed it. The battery's terminals are actually "members" of two capacitors. Take for instance the battery's positive terminal: it acts both as the plate of a capacitor having the Earth as other plate (battery's positive terminal/Earth capacitor), and as the plate of the capacitor having the battery's negative terminal as other plate ("intrinsic/internal battery's terminals capacitor).

The battery EMF can only "redistribute/rearrange" the charges between the "internal/intrinsic battery's capacitor plates", however it can't push charges on the plate of the battery's terminal/Earth capacitor (since no current can flow from the Earth through the battery because there is not a closed electric path) - see also - http://amasci.com/emotor/cap1.html

That depends on the previous history, not just the battery EMF. Previous manipulations could have left some overall charge on the battery.

As above, I believe battery EMF is not involved at all. Battery EMF can only "move" charges by a current flowing through it. Since there is not a closed circuit involving the battery and the Earth, there is no way for the battery EMF to move charge on it. Possibly charges stored on the battery's positive terminal must be "placed" there from external processes that do not involve the battery EMF.

 

[cianfa72]

A battery stores charges internally. The small terminal capacitance of any component is determined by the geometry of the terminals, and the placement of those terminals in space. The difference in terminal voltage is an internal property of the voltage source component. The common charge on the whole flying battery component is an external property, quite independent of the internal charge.

Sorry, why do you talk of common charge on the whole battery as a single flying component? As the battery flies around electrostatic fields in the space, each of the two terminals will have its own charge on it and its capacitance w.r.t. the Earth.

The conductive wire in a SPICE simulation, that connects common component terminals, is a single equipotential node of zero size. A node has no series resistance, nor capacitance to ground, unless you choose to include those in your model as discrete components. Currents are summed into a node as a single equipotential point.

The wire, as a node in SPICE, does not have inductive or transmission line properties, nor any current direction, unless you need to subdivide one node into several nodes, and include those subcomponents in your circuit model.

You wanted to model charge on the battery terminals, but that defeats the terminals being zero-size nodes, so we added the terminal capacitance to ground, just to keep you happy. Now you seem to be complaining that the terminal capacitance of a flying battery, has variable geometry, as it flies through undefined electrostatic space, to arrive at the circuit assembly point. SPICE does not model flying components during the assembly process.

Why ? That was not my point/complain.

 

[Baluncore]

Sorry, why do you talk of common charge on the whole battery as a single flying component? As the battery flies around electrostatic fields in the space, each of the two terminals will have its own charge and capacitance w.r.t. the Earth.

No.
If you try to electrostatically charge one terminal of the battery only, that charge will appear on both terminals in proportion to their effective capacitance. The difference in charge density between the terminals, will be related to the battery voltage, not which terminal you charged.

The external insulated case of the battery will also hold some residual surface charge that failed to leak across the surface to a terminal.

 

[cianfa72]

If you try to electrostatically charge one terminal of the battery only, that charge will appear on both terminals in proportion to their effective capacitance.

You mean the terminal own capacitance (just to be clear their self capacitance).

The difference in charge density between the terminals, will be related to the battery voltage, not which terminal you charged.

Ok, so the two battery's terminals will have a different charge on it (the total electrostatic charge will be redistributed according to the terminal self-capacitance and battery EMF).

The external insulated case of the battery will also hold some residual surface charge that failed to leak across the surface to a terminal.

Ok, this makes sense.

 

[Dale]

Since there is not a closed circuit involving the battery and the Earth

You can always make a closed circuit by including the capacitance with the ground.

 

[cianfa72]

You can always make a closed circuit by including the capacitance with the ground.

Ah ok, you mean the capacitors C1 and C2 in post#22 schematic when the battery is flying. Therefore, as you pointed out, the effective charge on battery's terminals depends on battery EMF and the "history" as it flies around the space.

As I said before, I'd add a capacitor C3 between the nodes Vn and Vp to model the "internal/intrinsic capacitance between terminals".

 

[Baluncore]

As I said before, I'd add a capacitor C3 between the nodes Vn and Vp to model the "internal/intrinsic capacitance between terminals".

That is pointless.
A voltage source has zero impedance. The Vp and Vn terminals are voltage clamped internally. Since the differential voltage cannot change, C3 will have a constant charge; C = Q / V = dq / dv .
Delete the idea of C3.

 

[Baluncore]

Let us assume that during the connection and grounding process, the excess electrostatic charge on the surface of the battery terminals is fixed.

As the flying battery falls towards the ground, the area, a, of the terminals remains constant, while the separation distance, d, between the terminals and ground, is being progressively reduced. The terminal to ground capacitance is proportional to, a / d, so the capacitance is increasing.

The definition of capacitance; c = q / v ;
v = Q / c .
Since the charge is fixed, and the value of capacitance is increasing, the battery terminal voltage will be falling relative to ground. At the instant of grounding, the voltage will reach zero, and the transported excess charge will be dumped, from the battery terminals to the ground.

 

[cianfa72]

Delete the idea of C3.

From a circuit perspective, sure, adding the capacitor C3 is pointless. However my point was to model how things actually work: the battery EMF will "move" electrons away from the terminal that will be the positive to the other terminal that will be the negative one. This process will keep going until the battery inter-terminal intrinsic capacitor will be fully charged at a voltage/PD equals to the battery EMF.

Let us assume that during the connection and grounding process, the excess electrostatic charge on the surface of the battery terminals is fixed.

Ok, as you pointed out before, the total electrostatic charge will be distributed on the surface of battery's terminals according to the relative effective terminal capacitances and battery EMF -- see Capacitance matrix

As the flying battery falls towards the ground, the area, a, of the terminals remains constant, while the separation distance, d, between the terminals and ground, is being progressively reduced. The terminal to ground capacitance is proportional to, a / d, so the capacitance is increasing.

The definition of capacitance; c = q / v ;
v = Q / c .
Since the charge is fixed, and the value of capacitance is increasing, the battery terminal voltage will be falling relative to ground.

Sorry, by the battery falling towards the ground, you actually mean the process by which the wire attached to the negative terminal is progressively brought closer and closer to the ground until the instant of grounding.

At the instant of grounding, the voltage will reach zero, and the transported excess charge will be dumped, from the battery terminals to the ground.

Ok, yes.

 

[Dale]

Delete the idea of C3

I am not so sure about that. Batteries have rate constants which could be meaningfully represented by an internal resistance and a capacitance. I am not sure if those would best be in series or parallel with the EMF.

 

[Baluncore]

I am not so sure about that. Batteries have rate constants which could be meaningfully represented by an internal resistance and a capacitance.

One or two picofarads of inter-terminal capacitance, is quite irrelevant to the model of a battery of chemical cells.

If you replace the battery with a model voltage source, any parallel capacitance becomes meaningless, as the capacitor voltage will never change.

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

 

[cianfa72]

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

What about an ideal voltage source V with a series resistor between the node Vn and Vp of post#22 schematic with a parallel C3 inter-terminal capacitance ?

 

[Baluncore]

What about an ideal voltage source V with a series resistor between the node Vn and Vp of post#22 schematic with a parallel C3 inter-terminal capacitance ?

As a battery, the series resistance would need to be less than 1Ω, or it would not be used. The time constant of the series resistance, with a 1pF terminal capacitance, would be of the order; 1Ω * 1pF = 1 picosecond. That is shorter than any simulation step.

For a 12 volt battery, the charge on the 1pF terminal capacitance would be;
q = c * v = 12 pC.
A 1 A⋅hr battery will store and deliver, 3600 C. The 12 pC terminal charge, will be lost in the round off of the 15th digit, during a double precision simulation.

 

[cianfa72]

Btw, I believe a sensitive approach, at least in principle, boils down to setting up a (quasi) Dirichlet problem for the Laplace equation. The boundary condition is the PD between battery's terminals (including the zero potential for the "Earth conductor"). As the battery flies through space, the system capacitance matrix changes accordingly, and battery EMF pushes/moves charges (i.e. electrons) on its terminals to keep fixed their PD voltage (in order to cope with the varying terminals capacitance matrix coefficients w.r.t. Earth).

In this model the charges on the 3 conductors involved (the two battery's terminals + Earth) no longer have to be equal even in magnitude.

 

[Dale]

Perhaps you could provide a model for a real battery, one that behaves differently when 2 pF of parallel inter-terminal capacitance is added.

I am not convinced that it will be 1-2 pF. For the internal capacitance the important structure would be the electrodes, which are usually much larger than the terminals.

One or two picofarads of inter-terminal capacitance, is quite irrelevant to the model of a battery of chemical cells.

You have staked your position on a number that I don’t accept.

The time constant of the series resistance, with a 1pF terminal capacitance, would be of the order; 1Ω * 1pF = 1 picosecond. That is shorter than any simulation step

Real batteries have measurable time constants that are longer than typical simulation steps. So one or both of these estimates is wrong.

Here is a 12 V Li battery model using two internal capacitances: https://www.nrel.gov/docs/fy01osti/28716.pdf I am not intending that as a definitive general model, just a single experimentally validated battery model that indeed shows capacitance.

 

[Baluncore]

Here is a 12 V Li battery model using two internal capacitances:

That models internal resistance as about 2 milliohm.
The short term chemical storage is modelled as 80 kilofarad.
Any Vp to Vn terminal capacitance, is swamped by the internal virtual voltage source.
Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

 

[Dale]

That models internal resistance as about 2 milliohm.
The short term chemical storage is modelled as 80 kilofarad.
Any Vp to Vn terminal capacitance, is swamped by the internal virtual voltage source.
Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

Sure. The small capacitance between the terminals and ground is different from the internal capacitance.

 

[cianfa72]

Any capacitance from Vp or Vn to ground, shares charge between the two and ground, without changing the battery voltage.

This because we have a loop consisting of voltage source and two capacitors (the capacitances between each terminal and ground). Technically the charges/voltages on such capacitors are not both state variables since there is a KVL constraint for the loop. Therefore the total charge is redistributed between the two terminal capacitance accordingly.

Indeed the above loop doesn't have a unique solution: one has to fix the charge on a capacitor (or their sum) in order to get an unique solution.

However, as I said in #41, I believe the above is just a rough model. This because there is no reason why the charges on the "Earth's plates" should be equal in magnitude to the charges on the respective battery terminal "plates".

 

[Baluncore]

Indeed the above loop doesn't have a unique solution: one has to fix the charge on a capacitor (or their sum) in order to get an unique solution.

The state variables are:
Q, excess charge on the flying battery.
C1, varying capacitance between ground and terminal p.
C2, varying capacitance between ground and terminal n.
Vp, voltage on p, relative to ground.
Vn, voltage on n, relative to ground.

Vb, the battery voltage, is a parameter. Vb = Vp - Vn .

Solve for Vn ;
Vp = Vn + Vb .
Q = ( Vp * C1 ) + ( Vn * C2 ) .
Vn = ( Q - Vb * C1 ) / ( C1 + C2 ) .

So the excess charge, Q, is shared between the terminal capacitors. That charge flows through the low-impedance voltage source during the balancing process.

Any inter-terminal capacitance, C3, between n and p is irrelevant, as the voltage is fixed by the low-impedance voltage source. V( C3 ) = Vb.

 

[cianfa72]

Solve for Vn ;
Vp = Vn + Vb .
Q = ( Vp * C1 ) + ( Vn * C2 ) .
Vn = ( Q - Vb * C1 ) / ( C1 + C2 ) .

 Ok. The excess charge you were talking about is just the total "external" electrostatic charge placed there or it also includes charges stored there by battery EMF ?

So the excess charge, Q, is shared between the terminal capacitors. That charge flows through the low-impedance voltage source during the balancing process.

By balancing process you mean the process that "redistribute" the excess total charge Q on battery's terminals/ground capacitor "plates".

 

[Baluncore]

The excess charge you were talking about is just the total "external" electrostatic charge placed there or it also includes charges stored there by battery EMF ?

The charges on the surface of the battery terminals, created by the internal chemical EMF, will appear between the terminals of the virtual C3 in this discussion. C3 is a single capacitor, so the charges on the terminals of C3 must be equal and opposite. The excess external charge, stored on C1 and C2, is independent of the charge on C3.

 

[sophiecentaur]

Ah ok, you did a simulation using LTspice. C1 and C2 are both discharged through R1 since the current through R1 is always the sum of I(C1) and I(C2).

Simulation programs can be a snare and delusion because they tend to make assumptions about those components in a circuit that you don't specify (or even notice). You should hesitate to conclude that the software 'got it wrong'; it just tried to help you by filling in the gaps in your data with its own estimates.

It's incredibly easy to find a disagreement between what your basic theory and a simulation tells you. So beware and choose the direction of what the basic theory tells you. When there's a problem then start looking for something that your basic approach has missed or failed to specify. EMFs can be difficult to quantify.

 

[cianfa72]

Simulation programs can be a snare and delusion because they tend to make assumptions about those components in a circuit that you don't specify (or even notice). You should hesitate to conclude that the software 'got it wrong'; it just tried to help you by filling in the gaps in your data with its own estimates.

With reference to the simulation in post#22, from my previous work, LTspice uses a Thevenin voltage source with a series resistor of 2 mOhm (or 500 Siemens Norton conductance) to drive the node referenced in the .IC V(nxx)=<value> global directive.

In this case the node Vp is driven with a Thevenin voltage source of 24V with a series resistor of 2 mOhm. With .tran simulations, LTspice evaluates first the Initial Transient solution (aka initial DC solution) opening the capacitor and shorting the inductors. Therefore in our case (before the grounding switch is turned on) the Initial DC solution for node Vn is 24 -12=12V w.r.t. ground. It is also the state initial condition for the numeric integration of the C2 capacitor voltage.

Said that I believe, as @Dale pointed out before, that even the battery EMF is "responsabile" for the charges stored on the battery's terminal/ground capacitors (i.e. C1 and C2 in the model). Consider the following: suppose no external charge is placed on the battery terminals and the inter-terminal capacitance is negligible. The PDs between each terminal and ground cannot be both zero. Hence the battery EMF must charge someway the battery terminal plates.