In electromagnetism, permittivity or absolute permittivity is the measure of resistance that is encountered when forming an electric field in a medium. In SI units, permittivity ε is measured in farads per meter (F/m or F·m−1);

My question is how the unit permitivity is defined, if unit permitivity is defined as 1F/1m what does it mean as 1Newton is defined as a force that gives acceleration of 1m/s2 to 1kg mass. What is the significance of Farad if one is concerned with electric field of a single positive or negative charge and not about capacitor field.

  • \$\begingroup\$ Maybe "resistance" shouldn't be the term that is used. Maybe "inertia" is better. \$\endgroup\$
    – Andy aka
    Dec 10 '16 at 9:48
  • \$\begingroup\$ This definition is quoted from wikipedia @Andyaka \$\endgroup\$ Dec 10 '16 at 9:52
  • 1
    \$\begingroup\$ When looking for the meaning of the units of a (material) constant, it's often most useful to look at and understand the formulae they are used in. That's how I learnt most of my EE formulae and constants when I was younger. I have to run now, so hopefully this hint will get someone else going. Or maybe you yourself. (Do write the answer here, I think it's a good question to keep around). \$\endgroup\$
    – Asmyldof
    Dec 10 '16 at 10:09
  • \$\begingroup\$ You don't talk about Farad if you are not concerned with Capacitance. \$\endgroup\$ Dec 10 '16 at 14:18

Let there be an infinite length parallel plate capacitor of area 1 square meter and the distance between them be 1 meter. Let the medium between them have a permittivity of 1 F/1m. It just says that, if one electron (1.6*10^-19 C) is present on both the parallel plates, then one needs to do 1eV (or 1.6*10^-19 Joules) of work to move an electron from one plate to another. The Presence of a single electron on each plate does have an electric field in between. And so you have to do work to move a charge from one end to another. And so you get capacitance.

C = Aε/d = Q/v

Note: The presence of the third electron is assumed to not to contribute to capacitance. Everything is just an approximation and mathematical modeling. Remember, even the electric field due to a parallel plate capacitor derivation just assumes that the plates are infinite and perfectly parallel. In the real word this is not the case (where we include the edge effects). We just model the NET effect by some assumptions with the help of maths.


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