. Three identical capacitors $C _1, C _2$ and $C _3$ have a capacitance of $1.0 \mu F$ each and they are uncharged initially. They are connected in a circuit as shown in the figure and $C _1$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{ r }$. The cell electromotive force (emf) $V_0=8 V$. 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 $C _3$ is found to be $5 \mu C$. The value of $\varepsilon_{ r }=$. . . . .
IIT 2018, Advanced
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Applying loop rule,
$\frac{5}{1}-\frac{3}{\varepsilon_r}-\frac{3}{1}=0$
$2-\frac{3}{\varepsilon_{ r }}=0$
$\varepsilon_{ r }=\frac{3}{2}=1.50$
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