Question

Three identical capacitors ${\text{C}}_1,{\text{C}}_2$ and ${\text{C}}_3$ have a capacitance of $1\mu \text{F}$ each, and they are uncharged initially. Then they are connected in a circuit as shown in the figure and ${\text{C}}_1$ is filled completely with a dielectric material of relative permittivity ${\epsilon }_r$. The cell's electromotive force (emf) is $V_0=8\text{V}$. First the switch ${\text{S}}_1$ is closed while the switch ${\text{S}}_2$ is kept open. When the capacitor ${\text{C}}_3$ is fully charged, ${\text{S}}_1$ is opened and ${\text{S}}_2$ is closed simultaneously.
When all the capacitors reach equilibrium state (steady condition), the charge on ${\text{C}}_3$ is found to be $5\mu \text{C}$. The value of ${\epsilon }_r$ is
(Upto one decimal places)

Answer 1

When ${\text{S}}_1$ is closed keeping ${\text{S}}_2$ open,


Charge stored in ${\text{C}}_3$:

$q=CV=1\times 8=8\mu \text{C}$.

As the switch ${\text{S}}_1$ is opened and ${\text{S}}_2$ being closed, capacitors ${\text{C}}_1{\text{C}}_3,{\text{C}}_2$ will form a closed loop & battery $V_0$ no longer remains part of circuit.


Some charge will be released from capacitor $C_3$ so that a common potential for $({\text{C}}_1,{\text{C}}_2)$ and ${\text{C}}_3$ is achieved.

Given, that in steady condition (equilibrium state) charge on ${\text{C}}_3$ is $5\mu \text{C}$.

Charge flowing through the circuit,
$q=8-5=3\mu \text{C}$

Circuit at equilibrium state:


Applying $KVL$ in closed loop, $(\Delta V=0)$

$(+\frac{5}{C_3})-\left(\frac{3}{{\epsilon }_r{C}_1}\right)-\frac{3}{C_2}=0$

As, $C_1=C_2=C_3=1\mu \text{F}$

$\Rightarrow \frac{5}{1}-\frac{3}{{\epsilon }_r}-3=0$

$\Rightarrow \frac{3}{{\epsilon }_0}=2$

$\therefore {\epsilon }_r=\frac{3}{2}=1.50$

Accepted answer: $1.5$