Electrostatics 6 Question 8
8. At $t=0$, switch $S$ is closed. The charge on the capacitor is varying with time as $Q=Q_{0}\left(1-e^{-\alpha t}\right)$. Obtain the value of $Q_{0}$ and $\alpha$ in the given circuit parameters.
$(2005,4 M)$
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Answer:
Correct Answer: 8. $Q_0=\frac{C V R_2}{R_1+R_2}, \alpha=\frac{R_1+R_2}{C R_1 R_2}$
Solution:
$Q_{0}$ is the steady state charge stored in the capacitor.
$Q_{0}=C[\mathrm{PD}$ across capacitor in steady state $]$
$=C$ [steady state current through $\left.R_{2}\right]\left(R_{2}\right)$
$$ \begin{aligned} & \quad=C \frac{V}{R_{1}+R_{2}} \cdot R_{2} \\ & \therefore \quad Q_{0}=\frac{C V R_{2}}{R_{1}+R_{2}} \\ & \alpha \text { is } 1 / \tau_{c} \text { or } \frac{1}{C R_{\text {net }}} \end{aligned} $$
Here, $R_{\text {net }}$ is equivalent resistance across capacitor after short circuiting the battery. Thus,
$$ \begin{aligned} R_{\text {net }} & =\frac{R_{1} R_{2}}{R_{1}+R_{2}} \quad\left(\text { As } R_{1} \text { and } R_{2} \text { are in parallel }\right) \\ \alpha & =\frac{1}{C \frac{R_{1} R_{2}}{R_{1}+R_{2}}}=\frac{R_{1}+R_{2}}{C R_{1} R_{2}} \end{aligned} $$
