Explanation

(c) The charge $q_{1} = CV_{0}$
or
$V_{0} = \frac{q_{1}}{C}$ …(i)
[image:2_image111.emf] $\therefore\ $Capacitors are in parallel, in parallel $V_{0}$ is same for all capacitors.
$\therefore$ For second capacitor $V_{0} = \frac{q_{2}}{2C}$ …(ii)
From Eqs. (i) and (ii),
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q_{2} = 2q_{1}$ …(iii)
After disconnecting the battery, the region between the plates of the capacitor $C$ is completely filled with a material of dielectric constant $(K = 2).$
Then, $\ V_{1} = \frac{q_{1}}{CK} = \frac{q_{1}}{2C}$
and $V_{1} = \frac{q_{2}}{2C} = \frac{2q_{1}}{2C} = \frac{q_{1}}{C}$ [from Eq. (iii)]
[image:2_image112.emf]
Charge will flow from 2 to 1 till
$\frac{q_{2}^{''''''''}}{2C} = \frac{{q''''''''}_{1}}{KC}$
$\frac{q_{2}^{''''''''}}{2C}\ = \frac{q_{1}^{''''''''}}{2C}\ \ \ \ $
$ie,\ \ q_{1}^{''''''''}\ \ = q_{2}^{''''''''}$
Earlier potential $\ \ \ \ \ V_{0}\ \ = \frac{q_{1}}{C}$
Now it is $V_{0}\ \ \ = \frac{q_{1}^{''''''''}}{2C}$
Now, $\ q_{1} + q_{2}\ \ \ \ = 3q_{1}$ [from Eq.(iii)]
and $q_{1}^{''''''''} + q_{2}^{''''''''} = \ 3q_{1}$
or $2q_{1}^{''''''''} = 3q_{1}\ or\ q_{1}^{''''''''} = \frac{3q_{1}}{2}$
$\therefore$ Now potential $\frac{q_{1}^{''''''''}}{2C} = \frac{3q_{1}}{4C}$
$V = \frac{3V_{0}}{4}$
[$\because q_{1} = V_{0}C$]