Re: [Phys-l] Capacitance problem

["Bob Sciamanda" <trebor@winbeam.com>]

Yes, indeed.  At the end of subsequent rounds, the charge on C1 is:
Q, 3/2 Q, 7/4 Q, 15/8 Q, 37/32 Q .......
This is the series:X[0]=0 and    X[n] = (1/2)X[n-1] + Q,  Where Q =CE/2, the 
first charge deposited on C1 in round 1..
This series converges to 2Q, so that V1 converges to E, as you surmised.

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
trebor@winbeam.com
http://www.winbeam.com/~trebor/
----- Original Message ----- 
From: "Brian Whatcott" <betwys1@sbcglobal.net>
To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
Sent: Saturday, March 29, 2008 7:40 PM
Subject: Re: [Phys-l] Capacitance problem


> Hmmm...you compute 0.5 V as the pd across the cap after the first cycle,
> and 0.75 V after the second.
> You might easily conclude there is a  pattern developing!
>
> :-)
>
> Brian
>
> At 03:05 PM 3/29/2008, you wrote:
> > Boo-boo !
> >
> > In the last two fiqures, it should read
> >
> > V1 = 3/4 E, not 3/4 CE
> >
> > Bob Sciamanda
>
>
>
> > > I understand that the original question was not addressed to the first
> > > cycle  of operation, - but to the result of numerous cycles.
> > >
> > >  Does it seem to you that the upper capacitor ends
> > > up with the battery potential across its plates?
> > >
> > >
> > >
> > > Brian Whatcott    Altus OK    Eureka!
>
>
> Brian Whatcott    Altus OK    Eureka!
>
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