91Ó°ÊÓ

We need to find $Q=Q_{max}\left(1-e^{-t/\mathrm{Ï„}}\right)$where$Q_{max}=C\mathrm{Î\mu }$.

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.

$∑V=0\left(1\right)$

The voltage across the battery is $\mathrm{Î\mu }$, 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:

$\mathrm{Î\mu }−\text{Δ}{V}_{\text{C}}−IR=0\left(2\right)$

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 $\mathrm{Î\mu }=\frac{Q\max }{C}$. We use the expressions for $\text{Δ}{V}_{\text{C}},I$and into equation $\left(2\right)$:

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

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

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

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 :

$Q=Q_{\max }\left(1−e^{−t/RC}\right)$