how can i prove that the charge in a charging capacitor in a DC circuit which has a battery, a resistor and a capacitor equals to: Q=Cε(1-e^(t/RC)

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how can i prove that the charge in a charging capacitor in a DC circuit which has a battery, a resistor and a capacitor equals to: Q=Cε(1-e^(t/RC)

Physics
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So you need a whole derivation? Regards, Electronics engineer.
as i write the potential difference in the circuit i come to a differential equation, but my anwer is not like the book's
do you know how to integrate or differentiate any complex thing?

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Start with Kirchoff's law for a closed circuit\[0=\epsilon-I(t)R-V_c(t)=\epsilon-\frac{dq}{dt}R-\frac q C\]where V_c is the voltage across the capacitor. A little rearrangement gives\[\frac{dq}{dt}=\frac{1}{R}(\epsilon-\frac{q}{C})\to\frac{dq}{(\epsilon-\frac{q}{C})}=\frac{1}{R}dt\to\frac{dq}{q-C\epsilon}=-\frac{1}{RC}dt\]integrate:\[\int_{0}^{Q}\frac{dq}{(q-C\epsilon)}=-\frac{1}{RC}\int_{0}^{t}dt\]\[\ln(\frac{Q-C\epsilon}{-C\epsilon})=-\frac{t}{RC}\to(\frac{C\epsilon-Q}{C\epsilon})=1-\frac{Q}{C\epsilon}=e^{-\frac{t}{RC}}\]\[Q(t)=C\epsilon(1-e^{-\frac{t}{RC}})\] \[\text{QED}\] :D
that was fun !
and you forgot that the exponent was negative Mojtaba
Why have I still not received a medal for this? I usually don't complain about that kind of thing, but this was a cool derivation! Hook it up people ;)

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