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Three identical capacitors $\mathrm{C}_1, \mathrm{C}_2$ and $\mathrm{C}_3$ have a capacitance of $1.0 \mu \mathrm{F}$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $\mathrm{C}_1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{\mathrm{r}}$. The cell electromotive force (emf) $V_0=8 \mathrm{~V}$. First the switch $S_1$ is closed while the switch $S_2$ is kept open. When the capacitor $C_3$ is fully charged, $S_1$ is opened and $S_2$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $\mathrm{C}_3$ is found to be $5 \mu \mathrm{C}$. The value of $\varepsilon_{\mathrm{r}}=$. . . .
(image)
$1.50$
$1.60$
$1.70$
$1.80$
Solution
(image)
Applying loop rule,
$\frac{5}{1}-\frac{3}{\varepsilon_{\mathrm{r}}}-\frac{3}{1}=0$
$2-\frac{3}{\varepsilon_{\mathrm{r}}}=0$
$\varepsilon_{\mathrm{r}}=\frac{3}{2}=1.50$
