How Does Charging a Capacitor Relate to Terminal Velocity in Physics?

[Chen]

When you charge a capacitor, the current through is given by:
$$I(t) = I_0e^{-\frac{t}{RC}}$$
Let's say that $$k = \frac{1}{RC}$$ so we can simply things to:

$$I(t) = I_0e^{-kt}$$

Theoretically, it should take the capacitor infinite time fully charge, at which point the current through it will be zero. But in reality it only takes a few seconds, a minute at most. This is because there's a minimal current that can go through the capacitor, which is defined as:
$$I_{min} = \frac{dq_{min}}{dt} = \frac{q_e}{1s} = 1.6x10^{-19} A$$
So we know that when the current I(t) hits that number, the charging is complete. We can also find the precise moment that the current stops.

My question is what about doing this for things like terminal velocity? When you jump from an airplane, how much time does it take to develop your terminal velocity? Your acceleration a(t) is also of the form:
$$a(t) = ge^{-kt}$$
To know when the capacitor is charged we used the minimum possible current, can we do the same here, i.e is there a minimum acceleration? Does it depend on Planck length maybe?

 

[mathman]

Terminal velocity of a falling object is a function of air resistance. That is why feathers and snow flakes fall so slowly. It has nothing to do with any Planck length.

 

[russ_watters]

Originally posted by Chen
My question is what about doing this for things like terminal velocity? When you jump from an airplane, how much time does it take to develop your terminal velocity? Your acceleration a(t) is also of the form:
$$a(t) = ge^{-kt}$$
To know when the capacitor is charged we used the minimum possible current, can we do the same here, i.e is there a minimum acceleration? Does it depend on Planck length maybe?

Its technically true that such a function takes an infinite amount of time to reach its equilibrium, but in the real, macroscopic world, that's a meaningless thing. There are too many other factors affecting your fall (air resistance as already said) to make that meaningful. It only takes about 10 seconds or so to reach terminal velocity jumping from a fixed object. From a plane, you're already at terminal velocity when you jump.

 

[Michael D. Sewell]

Russ is right, depending on the plane's airspeed, you are at or near terminal when you exit(the exit is usually the most exciting part of the jump). When skydiving, the air gets more dense as you approach the ground, so terminal velocity is lower. You are actually slowing down to terminal velocity during a long jump. In this respect, it is a slightly different problem than the theoretical capacitor problem.

I still remember how excited I was when I first discovered the connection between skydiving and capacitors; isn't physics fun?
-Mike

 

[Chen]

I'm afraid you are all missing the point.

Originally posted by russ_watters
Its technically true that such a function takes an infinite amount of time to reach its equilibrium, but in the real, macroscopic world, that's a meaningless thing.

So you agree that the acceleration function has a $$e^{-kt}$$ somewhere in it, even if the falling object is also affected by 100 other forces. So theoretically it would take the acceleration an infinite time to reach zero, because $$e^{-kt} = 0$$ when t tends to infinity. But that's obviously not true. My question is when exactly does the acceleration reach zero in real life? Obviously there is a minimum possible acceleration, just like there is a minimum possible current, there's no meaning to saying "A body is accelerating at $$10^{-100} \frac{m}{s^2}$$, is there?

 

[Chi Meson]

The "too many other factors" comment includes fluctuations in air currents, temperatures, pockets of pressure, not to mention subtle changes in orientation of the sky diver. So in real situations, acceleration continually rises and falls, going from various positives to various negatives. In other words, as you approach the mathematical, theoretical limit where air resistance equals weight, the situation becomes chaotic. THis is due to macro fluctiuations and is no where near anything quantum mechanical

 

[Chen]

Originally posted by Chi Meson
The "too many other factors" comment includes fluctuations in air currents, temperatures, pockets of pressure, not to mention subtle changes in orientation of the sky diver. So in real situations, acceleration continually rises and falls, going from various positives to various negatives. In other words, as you approach the mathematical, theoretical limit where air resistance equals weight, the situation becomes chaotic. THis is due to macro fluctiuations and is no where near anything quantum mechanical

You are completely missing the point of my question, rendering your answer meaningless. I do not really care what happens when you skydive.

 

[selfAdjoint]

Originally posted by Chen
You are completely missing the point of my question, rendering your answer meaningless. I do not really care what happens when you skydive.

No, you just want to push formulas to their unphysical limit. Even if something fell from very far away onto the surface of an airless planet, its final velocity would be finite, because it has only fallen a finite distance, in a finite time. And even if the universe is infinite you can't have something falling from infinitely far away because the age of the universe is finite (13.7 Gy, according to WMAP), and the greatest distance accessible is 13.7 X 10^9 light years. This is because according to relativity, everything beyond that is causally disconnected from us since no light signal can travel between us and there.


The bottom line is that, even though physics allows the concept of infinity, every physical process is represented as cut off at a finite level.

 

[Chi Meson]

Originally posted by Chen
You are completely missing the point of my question, rendering your answer meaningless. I do not really care what happens when you skydive.

No, I do get your point. OK, forget the skydiver. The "other factors" also holds for the charging capacitor. YOu are correct in saying that so small an acceleration is meaningless, and also a similarly small current is too variable to have meaning in the real world. THerefore, the question "when exactly does current/acceleration reach zero" is just as variable.

Consider a charging capacitor, once a few seconds have passed, and the theoretical current is down to 10^-12 amps or so. We are still well before considerations of the uncertainty priciple, but we have already passed our ability to measure it (or can we measure picoAmperes?). So in a sense, the current is essentially zero at the point where we can not measure any current.

So how many significant digits do you have in all the other measurements of the system? Say for example your initial current is measured (to the limit of precison of your meter) as 0.12345 amperes. Your current is for all PRACTICAL purposes zero as soon as it goes below 0.00001 amperes

 

[kuengb]

I think it's just a matter of being aware where the limits of a physical model are. If you just care about the technical, measurable current, a macroscopic differential equation will suffice. But then you cannot ask when I=0 will be reached, since if you do you are already thinking about single electrons (while your model assumes continuous charge) and you need a statistical model. Which will tell you, of course, that even if outer voltage is zero, there will probably be a current unequal to zero since electrons behave like a gas. And in the next step, you would use quantum stuff etc.

 

[da_willem]

Theoretically, it should take the capacitor infinite time fully charge, at which point the current through it will be zero. But in reality it only takes a few seconds, a minute at most. This is because there's a minimal current that can go through the capacitor, which is defined as: $$I_{min} = \frac{dq_{min}}{dt} = \frac{q_e}{1s} = 1.6x10^{-19} A$$

Why should this be a minimal current. E.g. when there is one electron per three seconds transported this should give rise to a current below the 'limit' you indicated. This 'minimal current' you mention is useful as a criterium to indicate the end of the charging of the capacitor, it's no physical limit.

So the fact that charge is quantized doesn't mean current is! And your argument certainly will not work in the case of accelleration because acceleration too is not quantized. So speaking of a under-limit is meaningles...

 

[Chen]

Good point da_willem, thank you.

 

[Michael D. Sewell]

Originally posted by Chi Meson
The "too many other factors" comment includes fluctuations in air currents, temperatures, pockets of pressure, not to mention subtle changes in orientation of the sky diver.

Often we make large changes in our orientation, and we approach terminal velocity from a much higher velocity. This is not totally irrelevant to your discussion of capacitors. In order to understand what is happening in a real life situation it is essential to understand the theories involved. It is equally essential to understand that in the real world there are often many (sometimes large) variables to contend with. Some say "Nature abhors a constant".

 

[Janitor]

Another technicality to throw a wrench in Chen's idea.

Doesn't any real-world charged capacitor have some leakage current across the dielectric? And also, when you take thermal effects into account, there will be some rather chaotic current flow in both directions (the charging direction of flow and also the discharging direction of flow) in the wires going to the capacitor plates whenever the capacitor is close to its full charge for the given applied voltage.

 

[Michael D. Sewell]

Janitor,
Exactly the point I was trying to make too. The system and the environment are changing all the time, and each changes the other. In the real world one does not get the results that are predicted with the constants, because the constants are not constant.

The theoretical skydive has a lot in common with the theoretical capacitor. The real world skydive has a lot in common with the real world capacitor as well.

One time in a lab we did Newtons law of cooling and someone turned up the thermostat and changed the room temperature on us.
-Mike