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 $\epsilon_{r}$. 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 $\epsilon_{r}=$ ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Capacitance

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} =

____.

Answer: 1.5

Initially with

S_{1}

closed,

C_{3}

charges to

q_{3} = C V_{0} = 8 μ C

. After

S_{2}

closes, charge redistributes. Using charge conservation and the condition that final charge on

C_{3}

5 μ C

\frac{8}{1 + \frac{ϵ _{r}}{1 + ϵ _{r}}} = 5

. Solving:

\frac{8}{5} = 1 + \frac{ϵ _{r}}{1 + ϵ _{r}}

, giving

ϵ_{r} = 1.50

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer