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There are two concepts I keep reading about that seem to be conflicting each other. Firstly, whenever there's a potential difference between two points, and those points are connected, electric current will flow between them, evening out the potential difference. And secondly, electricity needs a closed circuit to flow.

Consider a closed circuit that is floating (battery powered or isolated). If I were to connect the ground clip (connected to earth ground) on my oscilloscope probe anywhere in the circuit, it would be reasonable to assume that the electric potential at that point is different between that point, and earth ground.

Because the clip is connected to ground with a low resistance path, and because the potentials are different, it makes zero sense to me whenever people say that no current will flow through there. If current is the movement of electrical charges, how is it physically possible that connecting a different potential would not result in charge movement?

What I find even more confusing, is that people often quote how the earth has "infinite" capacity to dissipate charge, which does make sense, but then it would act as a capacitor with infinite capacitance? Or at least measurable non-zero capacitance, and since it is so large, there should be a non-trivial amount of current flowing to earth at least in the moment when the ground clip is connected.

On a similar note, if I we just consider a capacitor with no return connection as in this image

enter image description here

I find it very hard to believe that applying a potential difference to one side of the capacitor would not repel the charges on the other side, and even push out the charges at the end of the wire to act as a capacitor against its environment.

I do understand that in the case of a small capacitor this effect might be small enough for us to ignore, as the capacitance of a small wire is probably beyond what we can even measure with test equipment?

But in the case of connecting to earth ground, such as this

enter image description here

how come the "infinite capacitance" of earth ground doesn't cause the charges to flow there through a low resistance path? I would understand if the argument worked for something like batteries, where one could probably say that unless electrons flow in the battery it would not produce more electrons out of it (I'm not sure if this really is true), but in that case I would imagine there would be other power sources for which this is not true, such as if there was a capacitor on the circuit, and the earth ground connection would allow it to discharge itself into earth (because of potential difference) with low resistance quickly.

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  • \$\begingroup\$ Current flows in loops in man made circuits. So if you start by drawing the current path in a mesh from positive to negative battery terminal you will see where you err. Try drawing how u think the current is flowing through first. \$\endgroup\$
    – user1245
    Sep 25, 2019 at 19:14
  • \$\begingroup\$ how would you apply a potential difference to only one side of the capacitor? ... you cannot have a "difference" if you have only one item ... an apple cannot be different from itself \$\endgroup\$
    – jsotola
    Sep 26, 2019 at 0:57
  • \$\begingroup\$ @jsotola If the capacitor was discharged before being connected to the circuit both of its plates are at an equal potential. Once it is connected on one side to a high potential side of the circuit, chances are its own potential will be less (since it was discharged, so it has as much potential as a normal wire), and hence current would flow from the high potential area of the field (the power source) to the capacitor? \$\endgroup\$ Sep 26, 2019 at 1:35
  • \$\begingroup\$ "And secondly, electricity needs a closed circuit to flow." It depends. 1) In static electricity, no, it doesn't need it; an example is charging an object by induction. 2) In steady DC, yes, it does need it. 3) In AC, no, it doesn't need it; an example is a dipole antenna; another example is transmission lines. \$\endgroup\$
    – alejnavab
    Oct 23, 2020 at 6:15

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The problem at hand might be relating the physical world to the ideal circuit world.

There is no way for us to build this circuit (more on that later):

enter image description here

But if you could, there would be no charge flowing in or out of the ground connection.

In the real world we merely approximate things, we can connect a battery to a 1k resistor and connect a wire to ground, but the circuit really looks more like this:

schematic

simulate this circuit – Schematic created using CircuitLab

So there could be small amounts of current flowing to ground, the air can actually conduct a small amount of electrons, and there is a small amount of capacitance (and the whole circuit would function like an antenna of sorts.

We could even get more detailed than the model above!

The real question is, how detailed to you want your circuit model? The more complicated the circuit model, the more accurate it is, but the more time and energy is need to solve it.

Because the clip is connected to ground with a low resistance path, and because the potentials are different, it makes zero sense to me whenever people say that no current will flow through there. If current is the movement of electrical charges, how is it physically possible that connecting a different potential would not result in charge movement?

In a model like the one shown above, there is no current flow. In the real world there could be a small amount of current.

As far as your capacitor circuit goes:

I do understand that in the case of a small capacitor this effect might be small enough for us to ignore, as the capacitance of a small wire is probably beyond what we can even measure with test equipment?

The way this circuit is drawn, no current can flow through the capacitor, there is no other node for current to flow into. Both plates of the capacitor could be considered at the same voltage. Or the other node could be considered at an indeterminate voltage. (there is not a good way to calculate a differential voltage if you don't know the one of the values your differentiating).

In the real world, this is not the case, A real capacitor looks like this: enter image description here

The parallel resistance of the material between plates allows for a small amount of current that equalizes the voltage on the plates after a long time.

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    \$\begingroup\$ In the capacitor case, you say "there is no other node for current to flow into", but isn't that the same case as with earth ground, except that the wire at the end of the capacitor has much lower ability to absorb charge? I do understand that these values are probably small and practically insignificant, but I'm having a real hard time wrapping my head around "no current because not closed", and when do these things actually become large enough currents to possibly break something ... like when I touch a charged capacitor with my ground clip, or something like that \$\endgroup\$ Sep 25, 2019 at 18:02
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    \$\begingroup\$ The earth can sink very large currents, 1000's of Amps. A capacitor lead that is floating can sink 0 Amps. Earth can complete a circuit, meaning if you have two things that are plugged into earth ground current can flow through the earth like a wire \$\endgroup\$
    – Voltage Spike
    Sep 25, 2019 at 18:22
  • \$\begingroup\$ Sure but I don't mean the case when we close a circuit using ground. I mean a circuit where only one earth ground connection is made. Why is it that a floating capacitor lead can sink 0 amps, when you earlier said that everything has parasitic capacitance, including the other leg to the environment and rest of the circuit. \$\endgroup\$ Sep 26, 2019 at 1:38
  • \$\begingroup\$ It doesn't matter what circuit you have, you must have a closed connection or no current will flow. There must be a return path for the current \$\endgroup\$
    – Voltage Spike
    Sep 26, 2019 at 1:59
  • \$\begingroup\$ But why must I have a closed connection? If I were to take a loop of wire around a capacitor (to make them equal potential), cut the wire on both ends, charge the capacitor separately on a battery, disconnect it from the battery, and then connect one end of the first write to one end of the capacitor, there must be at least some current flow. If nothing else, the bunched up electrons in the capacitor that didn't have anywhere to go beforehand now had a lot more area to spread out in the wire ... and as a result, there would be current flow in the wire, no? \$\endgroup\$ Sep 26, 2019 at 2:14
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Firstly, whenever there's a potential difference between two points, and those points are connected, electric current will flow between them, evening out the potential difference.

This is true enough.

And secondly, electricity needs a closed circuit to flow.

This is an approximation, which is false specifically in the case you wish to consider. A better statement would be “electricity needs a closed circuit to continue flowing unchangingly”.

Whenever electric charges flow in a fashion that is not a closed loop, net charge must be being moved from one place to another (since the charges are not being somehow created or destroyed) — that means that there must be an accumulation of charge at one point. That accumulation of charge creates a net electric field, which opposes further accumulation of charge.

Accumulations of charge are electrostatic phenomena. We generally consider these separately from circuit analysis, except when within capacitors — because capacitors are devices intended to concentrate that phenomenon to a useful level for small voltages (as opposed to kilovolts).

Consider a closed circuit that is floating (battery powered or isolated). If I were to connect the ground clip (connected to earth ground) on my oscilloscope probe anywhere in the circuit, it would be reasonable to assume that the electric potential at that point is different between that point, and earth ground.

Yes, it probably is!

Because the clip is connected to ground with a low resistance path, and because the potentials are different, it makes zero sense to me whenever people say that no current will flow through there. If current is the movement of electrical charges, how is it physically possible that connecting a different potential would not result in charge movement?

Yes, charges will move, current will flow — very briefly, until the potentials are equalized. We call this event an electrostatic discharge (ESD), and since it may involve very high voltages (particularly if you shuffled across a carpet while carrying around the floating circuit, charging you and the circuit with respect to the surroundings) it is easy for such a discharge to cause damage to devices not robust against high voltage, if the discharge occurs to the wrong part of the circuit. Hence, ESD protection is an important part of electronic design.

I find it very hard to believe that applying a potential difference to one side of the capacitor would not repel the charges on the other side, and even push out the charges at the end of the wire to act as a capacitor against its environment.

It will! But this is a very, very small effect so we usually don't bother to consider it when studying circuits.

Skip thinking about ground, and about capacitors for the moment. Just consider two wires, connected to a battery and nothing else (not each other). Basic circuits tells us the battery creates a voltage, also known as a potential difference. The battery is acting to push charges “uphill” — there are more electrons in the negative wire than there would be otherwise, and fewer in the positive wire. If you then disconnected the wires from the battery, and touched them to each other, without discharging them by handling in the process (which is hard!), then a very small amount of charge would flow between them to bring them to equal potential.

(What we've done here is create a very poor capacitor. Just imagine bringing the wires closer together, or perhaps squishing them out flat into plates that we put an insulating sheet between, and you can see that there's no fundamental difference.)

When we want to model these electrostatic phenomena in terms of the concepts of circuit analysis, we can say that everything has some amount of capacitance between itself and every other object in the universe. But like gravity, that amount is usually too small to notice.

Outside of

  • the conditions that create static shocks and other noticeable and harmful ESD events, or
  • RF circuits or large AC circuits,

the energy that is stored and the current that flows as a result of these “stray” capacitances is so much smaller/briefer than the current that flows as a result of loops of conductors, that we like to leave it out of our analyses as a simplification, because most of the useful work comes from having a loop.

But in the case of connecting to earth ground, such as this [...] how come the "infinite capacitance" of earth ground doesn't cause the charges to flow there through a low resistance path?

“Infinite capacitance” isn't a good model (because an infinite capacitor is mathematically the same as a voltage source if charged or wire if not charged), but the answer is that charges do flow to equalize potential (on the side that is connected to ground). Once the potentials are equal, there is no reason for them to flow further along that path. Then the battery is free to make the potential on the other side of the circuit whatever it wants relative to earth.

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Even the voltage of Earth ground is a Floating Potential Voltage ; with respect to Mars or even TranOceanic cables which shunts hundreds of Amps from potential difference.

We always define ground as simply a 0V reference at one location. When the impedance and current is low enough when treated as noise "Vref=0" or Vn(f) = Z(f)*I(f) < "x" or effectively ~ 0V.

Then we can say that ground symbol or common grid or set of locations are effectively the same 0 V so we can still call it a ground. This applies regardless if it is floating with some leakage and the application can ignore noise below some threshold perhaps described in mV or maybe SNR (S/N)

Now any 3 pronged cordset with the Earth ground is actually a long wire that ends up connected to an earth grounding rod but has some inductance of ~ 1uH/m. But using the same formula above , we can say when connecting to that new ground (zero volt ref) we only have to consider the leakage impedance Z(f) for I(f)= V(f)/Z(f) or Ohm's Law spanning from DC to f max .

The resulting current between the two "grounds", battery 0Vdc & Earth Gnd , if any depends entirely on the leakage of the air gap between battery and earth ground.

So normally for DC we can assume for low voltage the air resistance is so high that the current may be neglected and thus no current will flow. But if the battery cables were wet and exposed to rain and dirt, we might be able to measure more leakage.

In 3 pronged unit with a SMPS, there is a line filter with ~ 1 to 4nf "Y caps" intentionally leaking RF noise to divert line noise to Earth Gnd. Smaller SMPS using 2 prong plugs dont bother with this filter, which often poses interference issues with communication because of leakage inside creating a cmoon mode yet high impedance voltage on both outputs. When connected to another high impedance circuit (loa), there is mor opportunity for Noise V(f) to be present, while having an earth bonds to shunt is has a low impedance can significantly improve results.

ESD example

Consider the air leakage moisture resistance from your body to a tribo-electric , nylon carpet. Your body acts as a large dielectric but without visible electrodes. If wearing plastic soled shoes, the plastic is another series capacitor and your feet/socks become one electrode while your high resistance finger can say be the other electrode.

When touching say a PC MOBO components even if disconnected from the grid or wall socket, there is still some leakage dielectric and resistance from the case to air to earth ground. Let's say you don't touch the case, but walk around on a nylon carpet (tribo-electric friction) and then zap a component on the floating PC MOBO how can this happen? The two electrodes are your finger air gap and socks. As the potential difference per mm exceeds the threshold for air (~1kV/mm or ~1V/um) the air can ionize making the high resistance air gap into a low resistance micro-arc and current can flow in less than a nanosecond to equalize the voltage. The current can easily exceed the 5 mA or so rating of CMOS reverse bias. ( |Vdd,Vss - Vzap| > 0.5) and the energy of this impulse determines if damage is done.

So the moist air here was the common potential and also the closed loop between the persons feet and finger becoming the electrode with the body dielectric building up the charge voltage with the breakdown voltage field of the closing air gap of the finger becoming like a gate-triggered SCR electrode.

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  • \$\begingroup\$ I think I understand most of the reasoning, except for why do my feet need to act as an electrode to ground, and why does there even need to be a loop for the ESD discharge? If I were to pick up the charges walking on a carpet, then float around to the PC and touch the board, while not being grounded with my feet, I would think that my finger would still zap the board with the same discharge, because of the accumulated excess charge that had nowhere to go until I got close to the board? \$\endgroup\$ Sep 26, 2019 at 1:49
  • \$\begingroup\$ Your shoes and feet make a big difference otherwise floating to there would be a much lower capacitance to earth with a large air gap. \$\endgroup\$ Sep 26, 2019 at 1:55

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