Within the capacitor,
$${E_1} = \frac{{{Q_1}}}{{2{\varepsilon _0}A}};{E_2} = \frac{{{Q_2}}}{{2{\varepsilon _0}A}};$$
where $$A$$ = area of each plate
$$d$$ = separation between two plate
$$E = {E_1} - {E_2} = \frac{1}{{2{\varepsilon _0}A}}\left( {{Q_1} - {Q_2}} \right)$$
Hence, $$V = Ed$$
$$ = \frac{1}{2}\frac{d}{{{\varepsilon _0}A}}\left( {{Q_1} - {Q_2}} \right) = \frac{{{Q_1} - {Q_2}}}{{2C}}$$

Capacitance are in series
$$\frac{1}{c} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \Rightarrow c = \frac{2}{3}F$$

$${C_1} = C,{V_1} = {V_0},{C_2} = KC,{V_2} = 0$$
$${\text{and}}\,{V_{{\text{common}}}} = V$$
We know that
$$\eqalign{ & {V_{{\text{common}}}} = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}} \cr & \therefore {V_{{\text{common}}}} = \frac{{C{V_0} + KC \times O}}{{C + KC}}\,{\text{or}}\,K = \frac{{{V_0}}}{V} - 1 \cr} $$

Given, $${C_1} = {C_2} = {C_3} = 4\mu F$$
(A) The network of three capacitors can be shown as

Here, $${C_1}$$ and $${C_2}$$ are in series and the combination of two is in parallel with $${C_3}.$$
$$\eqalign{ & {C_{{\text{net}}}} = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} + {C_3} = \left( {\frac{{4 \times 4}}{{4 + 4}}} \right) + 4 \cr & = 2 + 4 = 6\mu F \cr} $$
(B) The corresponding network is shown in figure below

Here, $${C_1}$$ and $${C_2}$$ are in parallel and this combination is in series with $${C_3}.$$
So, $${C_{{\text{net}}}} = \frac{{\left( {{C_1} + {C_2}} \right) \times {C_3}}}{{\left( {{C_1} + {C_2}} \right) + {C_3}}} = \frac{{\left( {4 + 4} \right) \times 4}}{{\left( {4 + 4} \right) + 4}}$$
$$ = \frac{{32}}{{12}} = \frac{8}{3}\mu F$$
(C) The corresponding network is shown below. All of three capacitors are in series.
So, $$\frac{1}{{{C_{{\text{net}}}}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$
$$\therefore C = \frac{4}{3}\mu F$$
(D) The corresponding network is shown below.

All of them are in parallel.
So, $${C_{{\text{net}}}} = {C_1} + {C_2} + {C_3}$$
$$ = 4 + 4 + 4 = 12\,\mu F$$
Thus, options (A) is correct.

$$\eqalign{ & {\text{Work done}}\,\, = \frac{1}{2}\frac{{{q^2}}}{C} = \frac{{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}}}{{2 \times 100 \times {{10}^{ - 6}}}} \cr & = 32 \times {10^{ - 32}}\;J \cr} $$

Given capacitor is equivalent to capacitors $${k_1},{k_2}$$  and $${k_3}$$ in parallel and part of $${k_4}$$ in series with them

$$\eqalign{ & \frac{1}{{{C_1}}} + \frac{3}{{{C_4}}} = \frac{{3d}}{{2{K_1}{\varepsilon _0}A}} + \frac{{3d}}{{2{K_4}{\varepsilon _0}A}} = \frac{{3d}}{{2{\varepsilon _0}A}}\left[ {\frac{1}{{{K_1}}} + \frac{1}{{{K_4}}}} \right] \cr & \Rightarrow {C_{{\text{eq}}}} = \frac{{K{\varepsilon _0}A}}{d} \cr & = \frac{{2{\varepsilon _0}A}}{{3d}}\left[ {\frac{{{K_1}{K_4}}}{{{K_1} + {K_4}}} + \frac{{{K_2}{K_4}}}{{{K_2} + {K_4}}} + \frac{{{K_3}{K_4}}}{{{K_3} + {K_4}}}} \right] \cr & K = \frac{2}{3}\left[ {\frac{{{K_1}{K_4}}}{{{K_1} + {K_4}}} + \frac{{{K_2}{K_4}}}{{{K_2} + {K_4}}} + \frac{{{K_3}{K_4}}}{{{K_3} + {K_4}}}} \right] \cr} $$

In series $$\frac{1}{C} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}$$    and charge on each capacitor is same.

$$\eqalign{ & {c_1} = \frac{{\left( {\frac{A}{2}} \right){\varepsilon _0}}}{{\frac{d}{2}}} = \frac{{A{\varepsilon _0}}}{d}, \cr & {c_2} = K\frac{{A{\varepsilon _0}}}{d},{c_3} = K\frac{{A{\varepsilon _0}}}{{2d}} \cr} $$

$$\therefore {c_{{\text{eq}}{\text{.}}}} = \frac{{{c_1} \times {c_2}}}{{{c_1} + {c_2}}} + {c_3} = \frac{{\left( {3 + K} \right)KA{\varepsilon _0}}}{{2d\left( {K + 1} \right)}}$$
($$\because {C_1}$$  and $${C_2}$$ are in series and resultant of these two in parallel with $${C_3}$$ )

Energy stored, in parallel plate capacitor is given by $$U = \frac{1}{2}\frac{{{q^2}}}{C}$$
$$\eqalign{ & {\text{but}}\,\sigma = \frac{q}{A}\,\,\,{\text{and}}\,\,C = \frac{{{\varepsilon _0}A}}{d} \cr & \therefore U = \frac{1}{2}\frac{{{{\left( {\sigma A} \right)}^2}}}{{\left( {\frac{{{\varepsilon _0}A}}{d}} \right)}} \cr & {\text{or}}\,\,U = \frac{{A{\sigma ^2}d}}{{2{\varepsilon _0}}} \cr & {\text{or}}\,\,U = \frac{1}{2}{\left( {\frac{\sigma }{{{\varepsilon _0}}}} \right)^2} \times {\varepsilon _0}Ad \cr & {\text{or}}\,\,U = \frac{1}{2}{E^2}{\varepsilon _0}Ad \cr} $$
Energy stored per unit volume i.e. energy density is thus given by
$$\eqalign{ & u = \frac{U}{V} = \frac{U}{{Ad}} = \frac{1}{2}{\varepsilon _0}{E^2} \cr & \Rightarrow u = \frac{1}{2}{\varepsilon _0}{\left( {\frac{V}{d}} \right)^2} = \frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}} \cr} $$

From the figure it is clear that the charge enclosed in the cubical surface is $$3C + 2C - 7C = - 2C.$$     Therefore the electric flux through the cube is