Three identical capacitors C_1, C_2 and C_3 h | vedclass

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(D) $1$. Initially,$S_1$ is closed and $S_2$ is open. $C_3$ is connected directly to the battery $V_0 = 8 \ V$. The charge on $C_3$ becomes $Q_3 = C_3

Vedclass · Accepted Answer

$1.50$

Vedclass · Answer

(D) $1$. Initially,$S_1$ is closed and $S_2$ is open. $C_3$ is connected directly to the battery $V_0 = 8 \ V$. The charge on $C_3$ becomes $Q_3 = C_3 V_0 = (1.0 \mu F)(8 \ V) = 8 \mu C$.$2$. When $S_1$ is opened and $S_2$ is closed,the charge $8 \mu C$ on $C_3$ redistributes among the capacitors $C_1, C_2$ and $C_3$. Let the final charge on $C_3$ be $Q_3' = 5 \mu C$. Since $C_3$ is in parallel with the series combination of $C_1$ and $C_2$,the voltage across $C_3$ must equal the sum of voltages across $C_1$ and $C_2$.$3$. The voltage across $C_3$ is $V_3 = \frac{Q_3'}{C_3} = \frac{5 \mu C}{1 \mu F} = 5 \ V$.$4$. The charge remaining for the series combination of $C_1$ and $C_2$ is $Q_{12} = Q_{initial} - Q_3' = 8 \mu C - 5 \mu C = 3 \mu C$. Thus,$Q_1 = Q_2 = 3 \mu C$.$5$. The voltage across $C_1$ is $V_1 = \frac{Q_1}{C_1'} = \frac{3 \mu C}{\varepsilon_r (1 \mu F)} = \frac{3}{\varepsilon_r} \ V$,where $C_1' = \varepsilon_r C_1$.$6$. The voltage across $C_2$ is $V_2 = \frac{Q_2}{C_2} = \frac{3 \mu C}{1 \mu F} = 3 \ V$.$7$. Applying the loop rule: $V_3 = V_1 + V_2 \implies 5 = \frac{3}{\varepsilon_r} + 3$.$8$. Solving for $\varepsilon_r$: $2 = \frac{3}{\varepsilon_r} \implies \varepsilon_r = \frac{3}{2} = 1.50$.

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