0

Voltage in a radial electric field is given by kQ/r (where k=1/(4*pi*epsilom.0)). Voltage in a circuit is given by IR, so if R=0 there is no change in voltage for an electron flowing through the circuit, (V=0). However in a vacuum, if you had just 2 (opposite) charges and no circuit, there would be no resistance but there would still be a change in voltage as the charges moved together. What is the difference between these 2 situations?

user294388
  • 49
  • The vacuum has infinite resistance, while an ideal wire has zero resistance. They are diametrically opposite physical situations. – CuriousOne Apr 13 '16 at 10:38
  • Thanks-is the 'resistance' of a vacuum related to the value of epsilom-nought? It's strange to think of a vacuum having resistance! – user294388 Apr 13 '16 at 16:01
  • A vacuum has nothing in that medium. There is nothing to conduct there. So vacuum has infinite resistance. – UKH Apr 13 '16 at 16:15
  • I would say that for the vacuum resistance is not really defined. You could also argue that the mean free path is infinite and according to drude theory the resistance is therefore 0. Other resistance definitions using $\text{Im}(\epsilon_r)$ will also yield 0 for the vacuum. If it was infinite how would you build a vacuum tube? – Jannick Apr 13 '16 at 16:51
  • @Jannick: Vacuum means that there are no charges in there. There is no conduction. – CuriousOne Apr 13 '16 at 18:37
  • @CuriousOne That is of course true. Therefore resistance is not well defined. Resistance is a measure of dissipation of directional kinetic energy into heat. In vacuum there is no such thing. Think about it. If you put a single electron into the vacuum what is the resistance now? – Jannick Apr 13 '16 at 19:05
  • @Jannick: Resistance doesn't depend on a conduction mechanism being available. If one isn't, then the numerical value is simply infinite, which agrees with measurements. – CuriousOne Apr 13 '16 at 19:31
  • @CuriousOne I guess we won't agree on this. I see your point that if you come from the $Z = dV/dI$ and you don't generate free charges as in a vacuum tube one could say $Z\rightarrow\infty$. But if you come from the dissipation or scattering time/length standpoint this just doesn't make sense. See also this old question – Jannick Apr 13 '16 at 20:43
  • @Jannick: There is nothing to disagree with. The measurements are crystal clear and so are the definitions. – CuriousOne Apr 13 '16 at 20:45
  • So what is the resistance of a vacuum tube then? – Jannick Apr 13 '16 at 21:14

1 Answers1

0

Potential difference is a more general term because the interaction is not specfified. It could be electromagnetic, gravitational... . Voltage is however specified as the potential difference of the electromagnetic interaction. Therefore voltage and potential difference are equivalent if we deal with the electromagnetic interaction only.

You also have to be careful with the understanding of ohm's law. For a finite resistance and voltage applied, electrons are accelerated by the electric field until the "friction" in the resistor counteracts the acceleration. If you reduce the resistance the velocity of the electrons will increase until equilibrium is reached again. The potential energy gained is converted to heat in the resistor due to this friction. For $R=0$ the electrons will acclerate more and more and the potential energy gain is converted to kinetic energy instead. This is what happens in a vacuum tube. Ohm's law only makes sense when electron motion can equilibrate in the resistor.

Jannick
  • 1,263
  • You are welcome to take a vacuum tube and put your Ohmmeter on it. What are you going to see? – CuriousOne Apr 13 '16 at 18:38
  • A finite non-zero value of course, but it is rather useless because a vacuum tube does not follow ohm's law. Above a certain threshold value the current should be constant and be equal to the number of free electrons generated. – Jannick Apr 13 '16 at 18:54
  • Ohm's law is just the linear version. $Z=dV/dI$ is useful at all times. For "vacuum", which in practice is really gas filled volumes, the value is very high, until avalanche ionization sets in and we are in gas discharge mode. For a perfect vacuum $dI=0$, which puts $Z$ at infinity. – CuriousOne Apr 13 '16 at 19:30