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I have been wondering really hard about what is the relationship between the defined zero potential at infinity as we do in physics, and the potential differences in circuits using batteries or other voltage sources.

Suppose we have a circuit with a battery and a resistor, and we measure the voltage across the resistor, what is the relationship between this potential and the zero defined at infinity?

As we know, the potential at a point due to an electric field is the integral from infinity to that point of E.dl, so how do you use this to calculate the potential across the resistor?

I know the zero of the circuit is in the circuit not at infinity as that would be dumb, but still this question hurts my brain. I want to know how do you calculate the potential across the resistor using the zero at infinity.

Can I define the potential across the resistor in relation to the zero at infinity or this doesn't exist ?

Can I calculate the potential across the resistor by taking a path integral from one end of it to the other ? Can I do it by taking the integral from one end of it to infinity, and then from infinity to the other end, and will the answer be the same as the voltage source reading?

Afterall, all the battery does is to split charges, so it creates an electric field, but I find it hard to visualize this electric field. I've heard that the electric field is zero across a conductor, but non zero in the resistor, so if I find the electric field inside the resistor to be constant, then the integral from one end of it to infinity and then back to the other end will be zero everywhere except inside the resistor if the electric field points directly from one end of the resistor to the other, but not if it doesn't.

What direction does the electric field inside this resistor point to?

Thanks a million!

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closed as unclear what you're asking by Voltage Spike, clabacchio Jul 18 '17 at 7:16

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    \$\begingroup\$ Look near the bottom of my answer here: electronics.stackexchange.com/questions/317018/… . You can see that I use a physicist definition of voltage there, complete with an equation that involves the difference between two integrals out to infinity to define voltage. A good book on the topic is Matter & Interactions. \$\endgroup\$ – jonk Jul 15 '17 at 3:47
  • \$\begingroup\$ @ jonk Thanks for recommending that book. Indeed its different from other E&M discussion. \$\endgroup\$ – analogsystemsrf Jul 15 '17 at 3:56
  • \$\begingroup\$ @analogsystemsrf I personally felt that it conveys better intuition, more quickly, lasting longer in mind, and with less effort than any other book I've read. \$\endgroup\$ – jonk Jul 15 '17 at 4:05
  • \$\begingroup\$ Jonk, I a reading your reply on the other post. I think there is a problem because if you increase the distance between the two plates charged to 10V, the voltage between them will increase. That's a fact about capacitors. Please advise further \$\endgroup\$ – Paulo Jul 15 '17 at 4:58
  • \$\begingroup\$ @Paulo I wasn't discussing capacitors. That was another person's comment. I did mention that the distance wasn't relevant to the energy released upon impact. But I wasn't talking about moving the plates with charge on them -- that's a different problem, entirely, which I hope you can see. I was just talking about two different experimental setups. Different problem, as I said. \$\endgroup\$ – jonk Jul 15 '17 at 7:29
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You're conflating two concepts, potential, and potential difference.

It's probably easiest to go right back to the definition of potential to see what's happening.

A potential field is one where the potential energy of an item that feels that field is dependent only on its position. For instance, in a gravity field, the energy of a test mass is given by its height alone. The change of energy in moving from height 1 to 2 is the same as that released by moving from 2 to 1. This energy is measured in Newton.meters (Nm) aka Joules.

In an electric field, you use a test charge instead of a test mass. It needs energy to move it to a higher potential, and releases the same amount of energy when moved to a lower potential. This energy is measured in charge.volts. In atomic physics, we use eV, electron.volts, but in electrical engineering, we tend to use coulomb.volts, aka Joules, aka watt.seconds.

What a multimeter does is allow us a short-cut way to estimate the energy changes that would happen as we move this test charge around. If we connect the probes across a battery, it will read directly what the voltage difference is between two points. If we took one coulomb of charge, and moved it from one terminal of a 12v battery to the other, it would absorb (if we were charging) or deliver (if the battery was discharging) 12 coulomb.volts, that is, 12 Joules.

Now let's push our test charge out from one battery terminal to a third point. The energy it took to do that is the potential difference of the third point to that at which we started. Now let's bring it back to the other terminal. The difference in energy for the whole trip will be 12 charge.volts. It doesn't matter (conceptually) where that third point is, at mains ground, on the moon, at infinity, the round trip journey still has the same change in energy.

As the only thing we can measure is the difference between two potentials, we can assign an arbitrary reference point. We could measure the potentials of several objects, then add the same constant to each measure, and the potential differences would stay the same.

For instance, in a gravity field, there are several common 'zero potential' reference levels. There's the floor level, local ground level, mean sea level, and deep space, depending on what your profession and purpose is.

In an electric field, electrical engineers tend to use earth potential or local chassis as zero, where physicists might sometimes prefer to use infinity. Either reference can be translated into the other by simply adding a constant to all potentials.

So 'zero potential at infinity' is a definition, not a measurement.

Let's see what potential to ground can look like under one measurement condition. This is a picture of a battery with 10pF capacity to ground (reasonable), a person with 100pF capacity to ground (reasonable, taken from the Human Body model for ESD testing), with 1pF capacity to the battery (equivalent to reaching out a hand, and holding it 100mm from the battery), and a battery represented by an internal capacity of 100kF (the battery voltage changes by 1 volt (12.8v down to 11.8v) when we pull 30Ah of charge from it, do the C=Q/dV sums).

schematic

simulate this circuit – Schematic created using CircuitLab

The first thing to notice is the huge dynamic range, 100kF/1pF = \$10^{17}\$.

The idea is that initially the battery was at ground potential, but then you walked across a carpet, getting charged to 11kV in the process, and then held out a hand and waved it 100mm from the battery.

You can see that 1pF capacity to the battery gets charged to about 10kV, moving a charge of CV = 10nC.

As that charge also charges the battery capacity to ground, the battery case gets charged to 1kV (Q/C = V). See how easy it is to alter the potential of things when they're not grounded.

That charge also flows through the battery (in the configuration I've drawn it). If we do V = Q/C again, we find the battery voltage has changed by 100fV (0.1pV), which is not very much.

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  • \$\begingroup\$ THanks for your indepth response. I understand that all we can measure are differences. What is hurting my mind at the moment is that it seems there is something special about batteries and that they don't respond to the same laws as static charges. If you consider the +ve terminal of a battery, it's full of +ve charge right? So does it create around itself an electric field, and does this electric field expand to infinity as well? And so if I take the integral from infinity to the +ve terminal will I get a +ve result? \$\endgroup\$ – Paulo Jul 15 '17 at 13:53
  • \$\begingroup\$ Another thing is, why is it that if I connect the -ve of one battery to the +ve of another, why isn't there a potential difference ? There's a lot of positive charge on +ve and negative on -ve, so won't they attract each other? There's no potential difference between these different battery terminals and this hurts my brain, because to me only the charges matter, but clearly there's something wrong here and I don't know what I am confusing but it's not potential difference and potential. \$\endgroup\$ – Paulo Jul 15 '17 at 13:57
  • \$\begingroup\$ The amount of charge depends on capacitance and voltage, and capacitance is a tricky function of electrode configuration. It's therefore difficult to deduce principles by trying to reason about volts and charge, unless you have a good basic understanding, which you are still striving for. Move a charge from infinity to the positive terminal, and infinity to the negative terminal, the difference in energy change will be test_ charge.battery_volts. If you connect the +ve of one bat to the -ve of another, they will be at the same voltage, you don't have voltage across a good conductor. \$\endgroup\$ – Neil_UK Jul 15 '17 at 14:50
  • \$\begingroup\$ So connecting the +ve of a battery to -ve of another puts them at the same potential and hence no current. I accept that if we use ideal conductors, but what about putting a resistor in between? Why won't current flow? If we connect +ve and -ve of the same battery clearly current flows, why is the situation different with a second battery if they are identical? Will any charge flow at all to achieve a balance or won't any charges flow. \$\endgroup\$ – Paulo Jul 15 '17 at 17:02
  • \$\begingroup\$ I imagine some +ve charge would flow from one battery to -ve of another, but then, as all a battery does is to split +ve and -ve charges, then the +ve side of the second battery is already full of +ve charges, and even if we dump +ve on its negative side to neutralize its -ve side, the battery can't send anymore +ve to its + side as it's already full, and so it won't work. This makes sense to me now. I guess this is why two batteries won't transfer charge between themselves with just 1 leg connected. \$\endgroup\$ – Paulo Jul 15 '17 at 17:02
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In the OP, you said (my emphasis)

Afterall, all the battery does is to split charges, so it creates an electric field, but I find it hard to visualize this electric field. I've heard that the electric field is zero across a conductor, but non zero in the resistor, so if I find the electric field inside the resistor to be constant, then the integral from one end of it to infinity and then back to the other end will be zero everywhere except inside the resistor if the electric field points directly from one end of the resistor to the other, but not if it doesn't.

'The integral' - there are several types of integral, things you can integrate, ways to do it. Unfortunately, my complex algebra was my weak suit at college, so I can't spell them all out or use accepted terminology.

The integral that's relevant here is the energy that's required to move a test charge from A to B. One way to do it would be the summation of the force.distance product (ie elements of work) as we move along a line from A to B, or \$\int_A^B force \,dl\$ . This integral is conservative, it doesn't matter what path is taken from A to B, the integral is always the same. This is the definition of a potential field.

A similar integral is \$\int_A^B field \,dl\$ . This gives us the voltage difference between those points instead of the energy difference. However, the difference between those two is the simple scalar scaling factor of the charge present on the test charge.

If you are getting different results for integrating from A to B either directly, or via a point at infinity, then you are using the wrong integral, or visualising potential incorrectly, or both.

When people say 'electric field', they usually mean the gradient of the potential, so volts/meter.

There is an electric field inside the resistor as well, that makes the charge carriers move (don't get hung up in quantum mechanics, classical will do at this stage).

Restart your mental model with a potential field that has a defined voltage everywhere, and where the line integral between two points is independent of the path taken (caveat, this doesn't work where magnetic fields are changing, looping round a transformer core does affect the line integral, but that's for a later lesson!) Then see the electric field as the gradient of the potential field.

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