The capacitor in $Figure28.38a$begins to charge after the switch closes at $t=0s.$. Analyze this circuit and show that ,$Q=Q_{max}\left(1-e^{-t/\tau }\right)$where$Q_{max}=C\epsilon$.

We need to find $Q=Q_{max}\left(1-e^{-t/\tau }\right)$where$Q_{max}=C\epsilon$.

We finding the loop rule where the loop rule is a statement the electrostatic force is conservative. suppose we go around a loop, measuring potential differences across circuit elements as we go and the algebraic sum of these differences is zero when we return to the starting point.

$\sum V=0\left(1\right)$

The voltage across the battery is $\epsilon$, across the capacitor is $\text{Δ}{V}_C$and across the resistor is $IR$Then, from the loop rule and if we travel clockwise, as:

$\epsilon -\text{Δ}{V}_{\text{C}}-IR=0\left(2\right)$

The capacitor is fully charged, the charges accumulate on the plates of the capacitor. This charge is related to the potential difference across the capacitor as:

$\text{Δ}{V}_{\text{C}}=\frac{Q}{C}$

The current is the change of the charge with times $I=\frac{dQ}{dt}$We are given the emf of the battery as $\epsilon =\frac{Q\max }{C}$. We use the expressions for $\text{Δ}{V}_{\text{C}},I$and into equation $\left(2\right)$:

role="math" localid="1650874831826" $\begin{array}{c}\epsilon -\text{Δ}{V}_{\text{C}}-IR=0\\ \frac{Q_{\max }}{C}-\frac{Q}{C}-\frac{dQ}{dt}=0\\ \frac{Q_{\max }-Q}{C}=\frac{dQ}{dt}\\ \frac{dQ}{Q_{\max }-Q}=\left(\frac{1}{RC}\right)dt\end{array}\left(3\right)$

We integrate both sides of equation ($3$) to get the next:

role="math" localid="1650875054844" $$\begin{gathered} {\int }_0^Q\frac{dQ}{Q_{\max }-Q}={\int }_0^t\left(\frac{1}{RC}\right)dt \\ -\ln \left(Q_{\max }-Q\right)+X=\frac{{\displaystyle t}}{{\displaystyle RC}}\left(4\right) \end{gathered}$$

Here, $X$is constant and at time $t=0$this constant is $\ln Q_{max.}$Use the expression of into equation $\left(4\right)$to get the next equation as:

role="math" localid="1650875100712" $\begin{array}{c}-\ln \left(Q_{\max }-Q^{\text{'}}+\ln \left(Q_{\max }\right)=\frac{t}{RC}\right.\\ \ln \left(\frac{Q_{\max }-Q}{Q_{\max }}\right)=-\frac{t}{CR}\\ \frac{Q_{\max }-Q}{Q_{\max }}=e^{-t/RC}\\ 1-\frac{Q}{Q_{\max }}=e^{t/RC}\end{array}\left(5\right)$

Solve equation ($5$) for to get as :

$Q=Q_{\max }\left(1-e^{-t/RC}\right)$