The charge flowing through $${C_4}$$ is
$${q_4} = {C_4} \times V = 4CV$$
The series combination of $${C_1},{C_2}$$  and $${C_3}$$ gives
$$\eqalign{ & \frac{1}{{{C_{{\text{eq}}}}}} = \frac{1}{C} + \frac{1}{{2C}} + \frac{1}{{3C}} \cr & = \frac{{6 + 3 + 2}}{{6C}} = \frac{{11}}{{6C}} \Rightarrow {C_{{\text{eq}}}} = \frac{{6C}}{{11}} \cr} $$
Charge flowing through capacitors $${C_1},{C_2}$$  and $${C_3}$$ will be same as they are in series.
So, $${q_1}$$ flowing through $${C_1},{C_2}$$  and $${C_3}$$ is given by
$${q_1} = {C_{{\text{eq}}}}V = \frac{{6C}}{{11}} \times V$$
Now, ratio of charge on $${C_4}$$ and $${C_2}$$ is given by,
$$\frac{{{q_4}}}{{{q_1}}} = \frac{{4CV \times 11}}{{6CV}} = \frac{{22}}{3}$$

$$C$$ and $$2C$$ are in parallel to each other.
$$\therefore $$ Resultant capacity $$ = \left( {2C + C} \right)$$
$${C_R} = 3C$$
Net potential = $$2V - V$$
$${V_R} = V$$
$$\therefore $$ Final energy = $$\frac{1}{2}{C_R}{\left( {VR} \right)^2} = \frac{1}{2}\left( {3C} \right){\left( V \right)^2} = \frac{3}{2}C{V^2}$$

When key is on the position $$BD,$$  the situation is shown in the figure.

When key is on the position $$AD,$$  the situation is shown in the figure.

The charge flows to each capacitor is $$36\,\mu C,$$  and so total charge flows through point $$O$$ is $$72\,\mu C.$$

For an $$AC$$  circuit containing capacitor only, the phase difference between current and voltage will be $$\frac{\pi }{2}\left( {{\text{i}}{\text{.e}}{\text{.}}\,{{90}^ \circ }} \right).$$
In this case current is ahead of voltage by $$\frac{\pi }{2}.$$
Hence, power in this case is given by $$P = VI\cos \phi $$     ($$\phi = $$ phase difference between voltage and current)
$$P = VI\cos {90^ \circ } = 0$$

$$\eqalign{ & 4\pi {\varepsilon _0}r = C \cr & \Rightarrow r = \frac{C}{{4\pi {\varepsilon _0}}} = \left( {1 \times {{10}^{ - 6}} \times 9 \times {{10}^9}} \right) \cr & = 9 \times {10^3}m \cr & = 9\,km \cr} $$

Charge on Capacitor, $${Q_i} = CV$$
After inserting dielectric of dielectric constant $$ = K$$
$${Q_f} = \left( {kC} \right)V$$
Induced charges on dielectric
$$\eqalign{ & {Q_{ind}} = {Q_f} - {Q_i} = KCV - CV \cr & = \left( {K - 1} \right)CV = \left( {\frac{5}{3} - 1} \right) \times 90pF \times 2V = 1.2nc \cr} $$

The given capacitance is equal to two capacitances connected in series where
$${C_1} = \frac{{{k_1}{ \in _0}A}}{{\frac{d}{3}}} = \frac{{3{k_1}{ \in _0}A}}{d}\quad \frac{{3 \times 3{ \in _0}A}}{d} = \frac{{9{ \in _0}A}}{d}$$
and
$${C_2} = \frac{{{k_2}{ \in _0}A}}{{\frac{{2d}}{3}}} = \frac{{3{k_2}{ \in _0}A}}{{2d}}\quad \frac{{3 \times 6{ \in _0}A}}{{2d}} = \frac{{9{ \in _0}A}}{d}$$
The equivalent capacitance $${C_{eq}}$$  is
$$\eqalign{ & \frac{1}{{{C_{{\text{eq}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} = \frac{d}{{9{ \in _0}A}} + \frac{d}{{9{ \in _0}A}} = \frac{{2d}}{{9{ \in _0}A}} \cr & \therefore {C_{eq}} = \frac{9}{2}\frac{{{ \in _0}A}}{d} = \frac{9}{2} \times 9pF = 40.5pF \cr} $$

As from the given diagram, potential difference across each capacitor is $$200\,V.$$
So, charge on each plate of capacitor is given by
$$\eqalign{ & Q = \pm CV\,\,\left[ {_{V = \,\,{\text{voltage or potential difference across the capactor}}}^{C = \,\,{\text{capacitance of capacitor}}}} \right] \cr & = \pm 25 \times {10^{ - 6}} \times 200 = \pm 5 \times {10^{ - 3}}C \cr} $$

By symmetry $${C_5} = 10\,\mu F$$
Capacitor between $$B$$ and $$C$$ can be removed
so, now $${C_{{\text{eq}}}} = \frac{{8 \times 8}}{{8 + 8}} + \frac{{8 \times 8}}{{8 + 8}} = 8\,\mu F$$

The energy stored in the capacitor $$ = \frac{1}{2}C{V^2}.$$
This energy will be converted into heat in the resistor.
∴ Heat produced = energy stored
$$\eqalign{ & = \frac{1}{2}C{V^2} \cr & = \frac{1}{2} \times 4 \times {10^{ - 6}} \times {\left( {400} \right)^2} \cr & = 0.32\,J \cr} $$