RC time Circuit; charge of capacitor as a function of time derivation

Hey guys! New to physicsforums. I wanted to ask a more conceptual question regarding RC time Circuits. I spent some time trying to derive the equations, and I feel like I'm not setting up the problem correctly. Here's my attempt:



Solutions according to profecssor:



1) ##q_{charge}(t)=C\epsilon(1-e^{-\frac{t}{\tau}})##

2) ##q_{discharge}(t)=q_{0} e^{\frac{-t}{\tau}}##



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my attempt at solution 1

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##q##-charge on capacitor

##R##-Resistance encountered in circuit

##C##-Capacitance of capaticor

##\epsilon##-Electromotive Force of battery

##t## -time since capacitor began charging

##\tau## - ##RC##

##i## -current





##C=\frac{q}{\epsilon}##

∴##q=C\epsilon##

##\epsilon=iR=\frac{dq}{dt}R##

∴##q=RC\frac{dq}{dt}=\tau\frac{dq}{dt}##

∴##\frac{1}{q}\frac{dq}{dt}=\frac{1}{\tau}##

∴##\int_0^t \frac{1}{q}\frac{dq}{dt} dt=\int_0^t \frac{1}{\tau} dt##

∴##\int_{q(0)}^{q(t)} \frac{dq}{q}=\int_0^t \frac{1}{\tau} dt##

∴##ln(q(t))-ln(q(0))=\frac{t}{\tau}##

∴##ln(\frac{q(t)}{q(0)})=\frac{t}{\tau}##

∴##e^{\frac{t}{\tau}}=\frac{q(t)}{q(0)}##

∴##q(t)=q(0)e^{\frac{t}{\tau}}##

So it's here that I'm like "I guess ##q(0)=c\epsilon## just for the sake of looking more like the professor's solution" <I realize how stupid this is... lol

∴##q(t)=C\epsilon e^{\frac{t}{\tau}}##



There must be something that I'm just totally missing here. The math seems right, so I'm thinking it's the set-up that I'm missing.



Any help is appreciated! Thank you!