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I have a mile long parallel plate capacitor AB. It may be taken to be two parallel conducting wire with rectangular cross sections and the wires are close to each other forming the capacitor. The capacitor is fully charged to a voltage V. It is then discharged at the end B through shorting it with a resistance R.

Question: What is the time taken for the capacitor to fully discharge?

What I have in mind is that when fully charged, the plates have uniform inner surface charge density +d/-d along the wires. While discharging, current flows along the wire. As it is well know that the drift velocity of current is extremely small - order of mm/sec - compared to one mile, I expect it would take ages for the excess electrons from the end A to finally travel to reach end B. So the time to fully discharge such a capacitor should be a rather large figure.

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    $\begingroup$ It is not necessary for a charge from the far end to make its way all the way to the near end. Only the discrepancy between the densities of positive and negative charges needs to equalize. This can be accomplished, for reasonable charge densities, by very small net motion of charge carriers. $\endgroup$ – Ben51 Jan 23 '18 at 16:01
  • $\begingroup$ I may disagree. In the positive plates, the +ve ions of the metal at end A have some atoms depleted of electrons. On the facing -ve plates, there are real electron excess at the end A; such real electrons must leave the end A. In fact, there are real flow of excess electrons along the -ve plates from ends A to B. $\endgroup$ – itsme Jan 23 '18 at 16:07
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    $\begingroup$ Absolutely the electrons must move. Point is they don't need to move far. $\endgroup$ – Ben51 Jan 23 '18 at 16:08
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    $\begingroup$ Others will comment or not as they choose. The electrons are slightly (very slightly) bunched up on one side and spread out on the other. They only need to shift a bit match their average spacing to that of the positive charges. So no, "Happy" does not need to go on a trip to B. $\endgroup$ – Ben51 Jan 23 '18 at 16:24
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    $\begingroup$ Your mile-long capacitor is a real thing, it is very well understood, and it has a name: It is called a transmission line. $\endgroup$ – Solomon Slow Jan 23 '18 at 16:47
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From "capacitor discharge": https://web.northeastern.edu/afeiguin/p1220-Fall2011/slides/chapter26-RC.pdf

The current i or charge Q decays exponentially - very quickly getting near zero. i(t) = i₀ exp(-t/RC); where i₀ = large initial current in R.

There is not any electron that makes a full 1 or 2 mile journey when the capacitor is discharging; the current flow is a movement of the "train of charge carriers". A typical electron drift speed may be 0.02 mm/s. With an electrons shift of a distance of 0.02mm, the charges crossing R may be sufficient to neutralize the original +Q charge in the upper plate fully discharging the capacitor.

Theoretically, capacitor discharge is independent of geometry; the 1 mile long factor may not change things.

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