Question

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 ${\in }_r$. The cell electromotive force (emf) $V_o=8V$. 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 $4\times {\in }_r$ is:

Answer 1

The capacitor $C_3$ charges to $8\mu C$ as the switch $S_1$ is closed.

Ref. image 1

But when switch $S_1$ is opened and $S_2$ is closed, then capacitor $C_3$ is charged to $5\mu C$, thus the net charge $8\mu C-5\mu C=3\mu C$ resides on $C_1$ and $C_2$, as they are connected in series.

Ref. image 2

So, Apply the Kirchhoff loop,

$\frac{q_1}{C_1}+\frac{q_2}{C_1{\epsilon }_r}=\frac{q_3}{C_3}$

$\frac{q}{C}(1+\frac{1}{{\epsilon }_r})=\frac{5}{1}$

$3(1+\frac{1}{{\epsilon }_r})=5$

${\epsilon }_r=1.5$