Let's say I hook two capacitor in a parallel circuit with a voltage source. Would the capacitor closer to the battery be charged up first? Because when the voltage excites and draws electrons from the plates, the electrons doesn't know where to go, so it goes through the first path it sees (i.e. the first capacitor) and it builds the plate of the first capacitor until it is done and then builds the next capacitor that follows?

Or would it start to build the first one first and then "evenly" builds charges on both capacitors because some of the electrons are actually getting repelled from the first plate of the first capacitor?


A good analogy for an electrical circuit is to model it by a system of pipes with water flowing through them. The water represents the electrons and the voltage represents the pressure in the pipes. In this model the capacitors would be modelled by header tanks i.e. the pressure can pump water up into the tanks until the pressure from the height of the water in the tanks balances the pressure in the pipes.

When you connect the voltage source this is like turning on the water pump. When you do this the pressure wave from the pump propagates through the water at the speed of sound, so the pressure wave will reach the nearer header tank first. However the speed of the pressure wave is much faster than the motion of the water, so even though the pressure wave reaches the nearer tank first this is a very small time difference relative to the time taken to fill the tanks.

To get back to your circuit: when you connect the voltage source the "voltage wave" travels at a bit less than the speed of light so the nearer capacitor will start to charge first. However, assuming the difference in the length of the wires is a few cm, say 3cm to make the maths easier, the nearer capacitor will start charging $10^{-10}$ seconds before the farther one. The electrons actually move rather slowly, see http://en.wikipedia.org/wiki/Drift_velocity for details, so the time taken to charge the capacitors is likely to be far far longer than $10^{-10}$ seconds. The end result is that both capacitors will appear to charge simultaneously.

  • $\begingroup$ +1 on the voltage wave, but I can't see how the header tank analogy explains the opposite charge on the other plate of the capacitor. $\endgroup$ – Emilio Pisanty May 1 '12 at 12:08
  • $\begingroup$ I suppose the analogy would be pumping from a tank below ground to a header tank, but I suspect this may be pushing the analogy to far :-) $\endgroup$ – John Rennie May 1 '12 at 12:30
  • $\begingroup$ The speed of the electrons is irrelevant, and it isn't particularly slow, it is about 5000km/s $\endgroup$ – Ron Maimon May 1 '12 at 13:28
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    $\begingroup$ 5000km/s isn't the drift velocity - that's the Fermi velocity. The drift velocity is relevant because it's proportional to the current and hence the rate of charging of the capacitors. $\endgroup$ – John Rennie May 1 '12 at 13:40
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    $\begingroup$ better hydraulic analogy: en.wikipedia.org/wiki/Capacitor#Hydraulic_analogy $\endgroup$ – Steve Byrnes May 1 '12 at 14:33

Don't get confused by the other answers which provide more detail than you want or need. The speed of light, speed of electrons, etc. is always irrelevant (in practice) for circuits. (It can be relevant for transmission lines.)

If you include the resistance of the wires, the capacitor which connects to the voltage via the less-resistive path will get charged faster. ("Path" here means path all the way around the circuit loop...A high resistance behind the capacitor is just as effective at slowing the charging as a high resistance in front of the capacitor.) If it was just a voltage and a capacitor with no resistance, the capacitor would "instantaneously" get charged to the appropriate voltage. But thanks to resistance, you can't charge the capacitor instantaneously, because you can't push an arbitrarily large current through the wire.

Keep in mind that the charge on the capacitors is an exponential that asymptotically approaches the final value. In theory neither of them ever quite get 100% charged, they get 99% charged then 99.9% then 99.99% etc.

Hydraulic analogy:

You can imagine the capacitors are rubber membranes like these in two parallel tubes, and you're pushing water towards these two tubes. But one tube has hair clogging it (high resistance). The membrane behind (or even in front of) the clog will stretch slowly as the water has to work its way through the hair. The membrane in the unobstructed tube will get pushed to near its maximally-stretched position much more quickly.

The length of the tube is not as directly relevant--except insofar as it affects resistance--because the water flows uniformly along the whole length of the tube. When you start pushing at one end, it "instantaneously" moves water everywhere along the whole tube. (The word "instantaneously" is an oversimplification, but unless the tube is many many miles long you will rarely need to worry about it!)


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