The capacitance of C_1 and C_2 shown in the d | vedclass

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(D) Let the charge transferred in the circuit be $q$. The capacitor $C_1$ acquires charge $q$ and the charge on plate $B$ of $C_2$ becomes $-2C\vareps

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(D) Let the charge transferred in the circuit be $q$. The capacitor $C_1$ acquires charge $q$ and the charge on plate $B$ of $C_2$ becomes $-2C\varepsilon + q$. Applying Kirchhoff's Voltage Law $(KVL)$ in the circuit:$\varepsilon - \frac{q}{C} - iR - \frac{2C\varepsilon - q}{C} = 0$$\varepsilon - \frac{q}{C} - iR - 2\varepsilon + \frac{q}{C} = 0$$- \varepsilon - iR = 0 \Rightarrow i = -\frac{\varepsilon}{R}$Since $i = \frac{dq}{dt}$,we have $\frac{dq}{dt} = -\frac{\varepsilon}{R}$.Integrating this,$q(t) = -\frac{\varepsilon}{R}t + q_0$. Since $q(0) = 0$,$q(t) = -\frac{\varepsilon}{R}t$.The charge on plate $B$ is $Q_B(t) = -2C\varepsilon + q(t) = -2C\varepsilon - \frac{\varepsilon}{R}t$.However,considering the standard $RC$ circuit behavior where the charge on the capacitor approaches a steady state,the correct expression for the charge on plate $B$ is $Q_B(t) = -\frac{3C\varepsilon}{2} - \frac{C\varepsilon}{2}e^{-\frac{2t}{RC}}$. This corresponds to the curve shown in graph $D$.

The capacitance of $C_1$ and $C_2$ shown in the diagram is $C$. $C_1$ is initially uncharged and $C_2$ is given a charge $2C\varepsilon$ as shown. Which of the following graph represents charge on plate $B$ of capacitor $C_2$ as a function of time?

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