Three identical capacitors C_1, C_2 and C_3 h | vedclass

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(A) $1$. Initially,$S_1$ is closed and $S_2$ is open. Capacitor $C_3$ is connected directly across the battery of emf $V_0 = 8 \,V$. $2$. The charge o

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$1.50$

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(A) $1$. Initially,$S_1$ is closed and $S_2$ is open. Capacitor $C_3$ is connected directly across the battery of emf $V_0 = 8 \,V$. $2$. The charge on $C_3$ becomes $Q_3 = C_3 V_0 = (1.0 \mu F)(8 \,V) = 8 \mu C$. $3$. When $S_1$ is opened and $S_2$ is closed,$C_3$ is connected in parallel with the series combination of $C_1$ (with dielectric $\varepsilon_r$) and $C_2$. $4$. Let the final charge on $C_3$ be $Q_3' = 5 \mu C$. The potential difference across $C_3$ is $V = \frac{Q_3'}{C_3} = \frac{5 \mu C}{1.0 \mu F} = 5 \,V$. $5$. The charge lost by $C_3$ is $8 \mu C - 5 \mu C = 3 \mu C$. This charge flows to the series combination of $C_1$ and $C_2$. $6$. The capacitance of $C_1$ with dielectric is $C_1' = \varepsilon_r C_1 = \varepsilon_r (1.0 \mu F)$. $7$. The equivalent capacitance of $C_1'$ and $C_2$ in series is $C_{eq} = \frac{C_1' C_2}{C_1' + C_2} = \frac{\varepsilon_r (1.0)}{\varepsilon_r + 1.0} \mu F$. $8$. The charge on this series combination is $3 \mu C$. Thus,$V = \frac{Q}{C_{eq}} \implies 5 \,V = \frac{3 \mu C}{\frac{\varepsilon_r}{\varepsilon_r + 1} \mu F}$. $9$. $5 = \frac{3(\varepsilon_r + 1)}{\varepsilon_r} \implies 5\varepsilon_r = 3\varepsilon_r + 3 \implies 2\varepsilon_r = 3 \implies \varepsilon_r = 1.50$.

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