Confused about Kirchhoff's Laws for RC Circuits - Discharging

[spoonerism]

Hi. When trying to derive the equation for voltage across a discharging capacitor in series with a resistor using Kirchhoff's laws, I got stuck. My attempt was that the voltage gain across the capacitor should equal the voltage drop across the resistor, therefore q(t)/C = i(t)*R, or q(t) - RC*q'(t) = 0. Solving this yields an equation similar to the actual equation, but it yields v(t) = V*e^(t/RC) when it should be V*e^(-t/RC). I have scoured the internet, and every other proof uses the KVL equation as q(t)/C + i(t)*R = 0, seemingly treating i(t)*R as a voltage rise instead of a drop. I am simply confused as to the logistics of this equation, and why it is + i(t)*R instead of - i(t)*R. Thank you for your help, in advance!

 

[Simon Bridge]

Checking against he physics - the current from the discharging capacitor should depend on the negative of the charge. So you've probably missed out a minus sign.

The loop law gives you: $$\frac{q(t)}{C}-i(t)R=0$$ ... where q(t) is the remaining charge on the capacitor plates. The physics for the capacitor tells you the $i(t)=-\dot q$ which you substitute into the loop equation.

 

[spoonerism]

Alright, that clears it up. Thanks a lot!

 

[Simon Bridge]

You can be forgiven for not realizing: the q here has a different context to how you usually think of it in connection with a current.

I had a look around and it seems that every single website kinda "glides" over that bit. i.e. "you only get a positive current when the charge on the cap is decreasing".

The only ones to include the step seem to be for more junior students and, even then, they don't exactly go out of their way to point it out. I think most students would just memorize the argument without questioning it so you are to be commended.

Generally:
Normally when you see something odd like that it means that the author has skipped a step ;)
Funny stuff with minus signs means you probably need to look closer at what any arrows mean.
Have fun.