Electrostatics Ques 131
- Three identical capacitors $C_{1}, C_{2}$ and $C_{3}$ 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 $C_{1}$ is then filled completely with a dielectric material of relative permittivity $\varepsilon_{r}$. 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 $S_{2} \quad$ is closed simultaneously. When all the capacitors reach equilibrium, the charge on $C_{3}$ is found to be $5 \mu \mathrm{C}$. The value of $\varepsilon_{r}=\ldots$
(2018 Adv.)
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Answer:
Correct Answer: 131.(1.50)
Solution:
Formula:
Distribution of Charges on Connecting Two Charged Capacitors:
Applying loop rule,
$$ \begin{aligned} \frac{5}{1}-\frac{3}{\varepsilon_{r}}-\frac{3}{1} & =0 \Rightarrow \frac{3}{\varepsilon_{r}}=2 \\ \varepsilon_{r} & =1.50 \end{aligned} $$
BETA
