Can't we use dq/dt in RC discharge?

[yosimba2000]

In the derivation of discharging RC circuit, you use I = -dq/dt to get Q = Q₀ e^-t/RC

But isn't dq/dt already implied to be negative, since charge on the capacitor is decreasing, and you force it to be negative using initial nd final conditions? So the final state is some low charge q, and the inital state is high charge Q.

Doing it this way gives Q = Q₀ e^t/RC

I can see that as t increases, the charge Q increases as well. It's the opposite of what I know it should be and I know it has to do with using I = -dq/dt, but I can't see WHY we use -dq/dt when dq/dt is already negative.

 

[cnh1995]

For a discharging capacitor,
I=I_oe^-t/RC.
This equation already takes care of the decaying nature of the current (and the charge).
Use I=dq/dt and integrate to get

Q = Q0 e-t/RC

 

[Ssnow]

Hi, the minus is due by the Kirchoff's voltage law, see https://en.wikipedia.org/wiki/Capacitor
Ssnow

 

[sophiecentaur]

Charge flow and charge flow per second (I) are both proportional to charge and in the same direction in the circuit (yes I know that is obvious but. . . .) For an equation of the form
A=A₀e^t/RC to apply , where A is some other quantity associated with the circuit, A would have to be inversely proportional to Q (or I), to account for the lack of a minus sign. (And apparent unlimited growth with time)
e^-c = 1/e^c
Just to re-state the obvious. So it has to be down to your Maths I think.