http://grephysics.net/ans/9677/62
ok .. once you read this question.. you see that it is clearly mentioned that they used resistive wires.. so i first concluded that once the charged capacitors are connected, they will discharge like a normal RC circuit and finally the potential will be 0..
but there is something entirely diff going on.. they have treated this prob as if the wires were ideal.. and had no resistance. since if there were resistance there would be dissipation of energy and also the concept of common voltage across both the capacitors would be wrong.
9677#62

 Posts: 58
 Joined: Sat Jun 14, 2008 10:19 pm
Re: 9677#62
I think the nature of wires does not matter when we are looking for final vlaues for this problem( after long time).Charge flows from the positive plate to the negative till they reach a common voltage. During this the current is nonzero causing some potential drop but once the plates arrive at a common potential, there will be no current and hence no voltage drop.
Conservation of energy ( electrical energy) does not hold here as some amount will be lost as heat, light etc.
one posible way:is to use conservation of charge:
(1 muF) ( 5) + ( 2muF) ( 5) = ( 1muF + 2MuF) V
V = 5/3 = 1.7 V
Conservation of energy ( electrical energy) does not hold here as some amount will be lost as heat, light etc.
one posible way:is to use conservation of charge:
(1 muF) ( 5) + ( 2muF) ( 5) = ( 1muF + 2MuF) V
V = 5/3 = 1.7 V

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: 9677#62
YES!!! now it makes sense.. great thanks!!!sravanskarri wrote:I think the nature of wires does not matter when we are looking for final vlaues for this problem( after long time).Charge flows from the positive plate to the negative till they reach a common voltage. During this the current is nonzero causing some potential drop but once the plates arrive at a common potential, there will be no current and hence no voltage drop.
Conservation of energy ( electrical energy) does not hold here as some amount will be lost as heat, light etc.
one posible way:is to use conservation of charge:
(1 muF) ( 5) + ( 2muF) ( 5) = ( 1muF + 2MuF) V
V = 5/3 = 1.7 V
One must keep in mind these two points before taking any ETS exam..
Rule1: ETS is never wrong.
Rule2: In case of any doubt refer to Rule 1.
by the way just for fun here is another question..
i discharge a charged capacitor across a resistor. as usual charge conservation will be there. but then where will the charges reside after a long time.. the initial energy stored will eventually be lost as heat from the resistor. and we know after a long time q in the capacitor will be 0. surely, microscopically the drift velocity of the electrons will eventually die out.. then what??
Last edited by blackcat007 on Mon Sep 07, 2009 10:33 am, edited 1 time in total.
Re: 9677#62
The wires/plates were initially neutral, we just redistributed the electrons when we charged the capacitor. Charges don't like being crammed together onto a capacitor plate. You need a battery or something similar to force them there. If you remove that, similar charges are going to do what they like to do; get as far away from each other as possible. They will distribute themselves evenly along the wires/plates (just how if you add charge to a conducting sphere it will distribute itself evenly along the surface). The end result will be the neutral wires/plates you started with. The microscopic classical picture here is that you have conduction electrons bouncing around off lattice sites, doing a random walk of sorts, but with no net electron drift.blackcat007 wrote:i discharge a charged capacitor across a resistor. as usual charge conservation will be there. but then where will the charges reside after a long time.. the initial energy stored will be eventually be lost as heat from the resistor. and we know after a long time q in the capacitor will be 0. surely, microscopically the drift velocity of the electrons will eventually die out.. then what??

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Re: 9677#62
another observation: if i connected the capacitors with the same charged plates together with ideal/resistive wires, then there won't be any flow of charge since they have a common potential of 5V across them.