Let me add that your loop equation (and its built-in conventions) should not
be restricted to any particular initial conditions. IE;, if properly
written they will apply even if your UPPER capacitor plate begins with the
positive charge. Once the actual initial conditions are applied, the
solution will reflect them in the resulting signs of the relevant quantities
(eg, q and i), in accordance with the defined conventions.
Bob Sciamanda
Physics, Edinboro Univ of PA (Em) http://www.winbeam.com/~trebor/
trebor@winbeam.com
----- Original Message -----
From: "Bob Sciamanda" <trebor@winbeam.com>
To: "PHYS-L Maillist" <phys-l@carnot.physics.buffalo.edu>
Sent: Sunday, February 19, 2006 10:01 AM
Subject: Re: [Phys-l] RC Disharge Analysis
| Serway says -q/R -IR = 0. Can anyone explain how that could be correct?
|
| This is correct if i is defined as positive when it is CW (in your
| previously drawn circuit), and q is the charge on the lower plate. Note
| this also defines the relation i = dq/dt
|
|| The only way I can get that is to draw the current through the resistor
in
| a
|| direction that would result in charging the capacitor rather than
|| discharging it.
|
| You are not drawing the actual current, you are merely defining the
meaning
| of a positive result for i.
|
| I agree that Serway should explain his conventions more specifically.
|
| Bob Sciamanda
| Physics, Edinboro Univ of PA (Em)
| http://www.winbeam.com/~trebor/
| trebor@winbeam.com