Explanation

(b) The circuit is given as
[image:2_image146.emf]
Let $q_{1}\ and\ q_{2}$ be the charge after switch $S$ has been closed.
Then, $V = \frac{q_{1}}{6C} = \frac{q_{2}}{3C}$
[image:2_image147.emf]
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{q_{1}}{2} = q_{2}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ q_{1} = 2q_{2}$ …(i)
But we know that, charge is conserved
$\ \ \ \ \ \ \ \ \ \ q_{1} + q_{2} = 3q + 6q$
or $\ \ \ \ q_{1} + q_{2} = 9q$ …(ii)
On putting the value of $q_{1}$Eq. (ii)
$\ \ \ \ \ \ \ \ 2q_{2} + q_{2} = 9q$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ 3q_{2} = 9q$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q_{2} = 3q$
Now, from Eq. (i)
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\ q}_{1} = 2 \times 3q$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ q_{1} = 6q$
Hence,$\ \ \ \ \ \ \ \ q_{1} = 6q,\ q_{2} = 3q$