Three identical capacitors , and have a capacitance of each and they are uncharged initially. They are connected in a circuit as shown in the figure and is then filled completely with a dielectric material of relative permittivity . [2018]

Question

Three identical capacitors $C_{1}$ , $C_{2}$ and $C_{3}$ have a capacitance of $1.0 μ F$ 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 $ε_{r}$ . 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 $5 μ C$ . The value of $ε_{r} =$ __________. [2018]

Smart Achievers · Accepted Answer

(1.5)

When switch $S_{1}$ is closed and $S_{2}$ is opened, capacitor $C_{3}$ becomes fully charged.

$∴$ $Charge on capacitor C_{3}, q_{3} = C_{3} V = 1 \times 8 μ C = 8 μ C$

When switch $S_{2}$ is closed and $S_{1}$ is opened, When all the capacitors reach equilibrium, charge on $C_{3}$ is found to be $5 μ C$ .

Therefore, charges on $C_{1}$ and $C_{2}$ are $3 μ C$ each.

Applying Kirchhoff's loop rule,

$\frac{C V_{0} - q}{C} - \frac{q}{ε_{r} C} - \frac{q}{C} = 0$

$\Rightarrow \frac{5}{1} - \frac{3}{ε_{r}} - \frac{3}{1} = 0$

$∴$ $5 = 3 (1 + \frac{1}{ε_{r}})$

$∴$ $\frac{1}{ε_{r}} = \frac{5}{3} - 1 = \frac{2}{3}$

$∴$ $ε_{r} = 1.5$

Solution

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