Consider a basic circuit where there's a battery and an infinite ohm resistor. The voltage across the resistor is the voltage of the battery level. However, there's a direct equivalency between voltage and charge. In an ideal model of a circuit there's no capacitance, so no charge on either end of the resistor. How can there be a voltage level across the resistor then?
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1$\begingroup$ Ideal models are always wrong, but they are often useful. A large part of understanding physics is knowing how to avoid drawing false conclusions from ideal models. $\endgroup$– John DotyCommented Jul 23, 2022 at 11:37
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2 Answers
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To a surprisingly good approximation the conduction electrons in a conductor behave like a fluid. This is known as the free electron model. This fluid can be compressed, and indeed this is exactly what we do when we charge a conductor. If for example you negatively charge a spherical conductor that means you are packing more electrons into the same volume i.e. you are compressing the "electron fluid".
This fluid is almost incompressible i.e. increasing its density by a tiny amount requires a lot of work. To return to our example of charging a spherical conductor, even for potentials of millions of volts we are making only a tiny change to the overall density of the conduction electrons. The change is typically too small to be measured, but it does exist.
Now consider the system you describe. When you connect your battery to the wires the chemical reaction in the battery battery pushes electrons out of the negative terminal into the wire so the conduction electrons in the wire are slightly compressed creating a very small but non-zero negative charge in the wire. Likewise electrons flow out of the other wire into the positive terminal of the battery to give the other wire a very small but non-zero positive charge.
And these density differences create a potential difference between the far ends of the two wires where they are connected to the infinite resistance. Presumably we could use an open switch for your infinite resistance, in which case we get the potential difference across the switch.
You say:
In an ideal model of a circuit there's no capacitance
but what I've described is exactly what capacitance is. The wires connected to the battery do have a capacitance because electrons can flow into and out of them is response to the battery potential. The capacitance is vanishingly small, so for a typical battery voltage the charge is vanishingly small, but the capacitance is not zero.
answered Jul 23, 2022 at 7:29
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$\begingroup$ Thanks, but I still have one question about your explanation though which isn't making sense to me. Assuming the distance of the wire between battery to open switch was really short, all the contribution to the voltage would be basically from a small section of the wire. Since you said the charge cannot be compressed much, this means that there's in fact very little charge on the wire. So how can one get a voltage of say, 100 across an open switch then? $\endgroup$ Commented Jul 23, 2022 at 17:06
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In an ideal model of a circuit there's no capacitance . . . . .
There have been very many question which illustrate a supposed paradox when the electrical circuit is made up of an incorrect number of lumped elements.
This is a good illustration of such a question as in your example you have stated that the circuit consists of two lumped elements, a battery and an infinite resistor.
A battery has two terminals made of a conducting material which in this case are separated by an ideal dielectric (the infinite resistance resistor).
Hence this circuit consists of an ideal battery and an ideal capacitor.
The electrochemical process within the battery moves charges so that one terminal becomes positive and the other terminal becomes negative.
Those charges then set up an electric field in opposition to the movement of the charges until eventual the electric field is large enough to stop any further movement.
There is now a steady potential difference across the terminals which is equal to the emf of the battery.
The two terminal capacitor is charged to that potential difference.
So your circuit actually has a battery, a resistance with infinite resistance and a capacitor.
Suppose now that the battery was part of a series circuit also consisting of a switch, connecting wires with no resistance, and a resistor.
With the switch in the off position it acts as a capacitor and there is a potential difference set up across the switch just as described as an answer to the original question.
As the switch is closed the capacitance of the switch increases and charge flows in the circuit before the switch is actually closed.
At the same time because the circuit has inductance, as the circuit is in the form of a loop, the charging rate is reduced due to the back emf induced in the circuit (Faraday).
The fact that inductance is present also means that after the switch is closed the current in the circuit does not reach its final (steady-state) value instantaneously.
In many simple cases the capacitances and inductances involved, which are often called parasitic, are so very small that their effect produces changes in the circuit over such small periods of time that they are not measured and therefore they are neglected.
