Consider the given figure,
When the switch $$S$$ is connected to point 1, then initial energy stored in the capacitor can be given as $$ = \frac{1}{2}\left( {2\mu F} \right) \times {V^2}.$$
When the switch $$S$$ is connected to point 2, energy dissipated on connection across $${8\mu F}$$  will be
$$\eqalign{ & = \frac{1}{2}\left( {\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}} \right).{V^2} \cr & = \frac{1}{2} \times \frac{{2\mu F \times 8\mu F}}{{10\mu F}} \times {V^2} \cr & = \frac{1}{2} \times \left( {1.6\mu F} \right) \times {V^2} \cr} $$
Therefore, $$\% $$ loss of energy $$ = \frac{{1.6}}{2} \times 100 = 80\% .$$

The network is equivalent to

Therefore equivalent capacitance
$$\eqalign{ & = \left[ {2C\,{\text{series}}\,C} \right]//\left[ {C\,{\text{series}}\,2C} \right] \cr & = 2\left( {\frac{{2C \times C}}{{2C + C}}} \right) \cr & = \frac{{4C}}{3} \cr} $$

Applying conservation of energy,
$$\frac{1}{2}C{V^2} = m.s\Delta T;\,\,\,\,\,V = \sqrt {\frac{{2m.s.\Delta T}}{C}} $$

According to Coulomb's law, force between two charges is directly proportional to product of charges and inversely proportional to square of distance between them. Thus,
$$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}{q_2}}}{{{r^2}}}\,\,......\left( {\text{i}} \right)$$
where, $$\frac{1}{{4\pi {\varepsilon _0}}} = $$   proportionality constant.
If a dielectric medium of constant $$K$$ is placed between them, then new force between them,
$${F_1} = \frac{1}{{4\pi {\varepsilon _0}K}} \cdot \frac{{{q_1}{q_2}}}{{{r^2}}}\,\,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we have
$$\eqalign{ & \frac{{{F_1}}}{F} = \frac{1}{K} \cr & {\text{or}}\,\,F' = \frac{F}{K} \cr} $$
Thus, new force decreases $$K$$ times.

When a capacitor is charged by connecting a battery across its plates, the initial energy stored,
$$U = \frac{{{q^2}}}{{2C}}$$
When the battery is disconnected, then the charge remains constant i.e. $$q =$$  constant. Now, another identical capacitor is connected across it i.e. the capacitors are connected in parallel, so the equivalent capacitance
$$\eqalign{ & {C_{{\text{eq}}}} = {C_1} + {C_2} \cr & = C + C = 2C \cr} $$
Thus, final energy stored by the system of capacitors,
$$\eqalign{ & U' = \frac{{{q^2}}}{{2{C_{{\text{eq}}}}}} \cr & U' = \frac{{{q^2}}}{{2\left( {2C} \right)}} \cr & U' = \frac{1}{2}U \cr & \therefore U' = \frac{U}{2} \cr} $$

The molecules of liquid will convert into induced dipole, get oriented along the electric field produced between the plates and rise due to force of attraction.

The given circuit can be redrawn as,

It is a balanced Wheatstone’s bridge $$\left( {{\text{as}}\,\frac{{{C_{AB}}}}{{{C_{BD}}}} = \frac{{{C_{AC}}}}{{{C_{CD}}}} = \frac{6}{6}} \right).$$
So, potential of $$B$$ and $$C$$ are equal and a $$6\mu F$$  capacitor between $$B$$ and $$C$$ is ineffective. The simplified circuit is shown as below.

Capacitors of $$6\mu F$$  and $$6\mu F$$  in upper arms are in series order, so
$${{C'}_{{\text{e}}{{\text{q}}_{\text{1}}}}} = \frac{{6 \times 6}}{{6 + 6}} = \frac{{36}}{{12}} = 3\mu F$$
Similarly, $$6\mu F$$  and $$6\mu F$$  in lower arms are in series order, so
$${{C''}_{{\text{e}}{{\text{q}}_{\text{2}}}}} = \frac{{6 \times 6}}{{6 + 6}} = 3\mu F$$
Now, $${{C'}_{{\text{e}}{{\text{q}}_{\text{1}}}}}$$  and $${{C''}_{{\text{e}}{{\text{q}}_{\text{2}}}}}$$  are in parallel, hence
$$\eqalign{ & {C_{{\text{net}}}} = {{C'}_{{\text{e}}{{\text{q}}_{\text{1}}}}} + {{C''}_{{\text{e}}{{\text{q}}_{\text{2}}}}} \cr & = 3 + 3 = 6\mu F \cr} $$

$${V_C} = \frac{{{Q_1} + {Q_2}}}{{{C_1} + {C_2}}} = \frac{{CV + 2VC}}{{KC + 2C}} = \frac{{3V}}{{K + 2}}$$

The total charge on plate $$A$$ will be $$ - 80\mu C.$$   If $${q_B}$$ and $${q_C}$$ be the charges on plate $$B$$ and $$C$$ then
$${q_B} + {q_C} = 80\mu C\,......\left( 1 \right)$$

Also $$2\mu F$$  and $$3\mu F$$  capacitors are in parallel. Therefore,
$$\eqalign{ & \frac{{{q_B}}}{2} = \frac{{{q_C}}}{3} \cr & \therefore \frac{{80 - {q_C}}}{2} = \frac{{{q_C}}}{3} \cr & \therefore 240 - 3{q_C} = 2{q_C} \cr} $$
This charge will obviously be positive.

When S and 1 are connected
The $$2\mu F$$  capacitor gets charged. The potential difference across its plates will be $$V.$$
The potential energy stored in 2 $$\mu F$$ capacitor
$${U_i} = \frac{1}{2}C{V^2} = \frac{1}{2} \times 2 \times {V^2} = {V^2}$$
When S and 2 are connected
The $$8\mu F$$  capacitor also gets charged. During this charging process current flows in the wire and some amount of energy is dissipated as heat. The energy loss is
$$\eqalign{ & \Delta U = \frac{1}{2}\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}{\left( {{V_1} - {V_2}} \right)^2} \cr & {\text{Here,}}\,{C_1} = 2\mu F,{C_2}8\mu F,{V_1} = V,{V_2} = 0 \cr & \therefore \Delta U = \frac{1}{2} \times \frac{{2 \times 8}}{{2 + 8}}{\left( {V - 0} \right)^2} = \frac{4}{5}{V^2} \cr} $$
The percentage of the energy dissipated $$ = \frac{{\Delta U}}{{Ui}} \times 100$$
$$ = \frac{{\frac{4}{5}{V^2}}}{{{V^2}}} \times 100 = 80\% $$