The two capacitors are in parallel so
$$C = \frac{{{\varepsilon _0}A}}{{t \times 2}}\left( {{k_1} + {k_2}} \right)$$

$$\eqalign{ & {C_1} < {C_2} \cr & \therefore \frac{{{C_1}}}{{{C_1} + {C_2}}} < \frac{1}{2}\,{\text{and}}\,\frac{{{C_2}}}{{{C_1} + {C_2}}} > \frac{1}{2} \cr & C = \frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}} = {C_1}.\frac{{{C_2}}}{{{C_1} + {C_2}}} > \frac{{{C_1}}}{2} \cr & {\text{Similarly,}}\,C < \frac{{{C_2}}}{2} \cr} $$

If the battery is removed after charging, then the charge stored in the capacitor remains constant.
$$q = {\text{constant}}$$
Change in capacitance,
$$C' = \frac{{{\varepsilon _0}A}}{{d'}}$$
As, $$d' > d$$
Hence, $$C' < C$$
As, potential difference between the plates of capacitor is given by $$V = \frac{q}{C}$$
So, for the new capacitor formed
$$V' \propto \frac{1}{{C'}}\,\,\left[ {q = {\text{constant}}} \right]$$
As capacitance decreases, so potential difference increases.
NOTE
If the battery remains connected, the charge stored increases. Also, the potential difference $$V$$ becomes constant.

When $$S$$ is closed, there will be no shifting of negative charge from plate $$A$$ to $$B$$ as the charge $$- q$$  is held by the charge $$+ q.$$  Neither there will be any shifting of charge from $$B$$ to $$A.$$

In steady state, flow fo current through capacitor will be zero.
Current through the circuit,
$$i = \frac{E}{{r + {r_2}}}$$

Potential difference through capacitor
$$\eqalign{ & {V_c} = \frac{Q}{C} = E - ir = E - \left( {\frac{E}{{r + {r_2}}}} \right)r \cr & \therefore Q = CE\frac{{{r_2}}}{{r + {r_2}}} \cr} $$

Charge on $${C_1}$$ is $${q_1} = \left[ {\left( {\frac{{12}}{{4 + 12}}} \right) \times 8} \right] \times 4 = 24\mu c$$
The voltage across $${C_p}$$ is $${V_P} = \frac{4}{{4 + 12}} \times 8 = 2v$$
$$\therefore $$ Voltage across $$9\mu F$$  is also $$2V$$
$$\therefore $$ Charge on $$9\mu F$$  capacitor $$ = 9 \times 2 = 18\mu C$$
$$\therefore $$ Total charge on $$4\mu F$$  and $$9\mu F = 42\mu c$$
$$\therefore E = \frac{{KQ}}{{{r^2}}} = 9 \times {10^9} \times \frac{{42 \times {{10}^{ - 6}}}}{{30 \times 30}} = 420\,N{c^{ - 1}}$$

For potential to be made zero, after connection
$$\eqalign{ & 120\,{C_1} = 200{C_2}\,\,\,\left[ {\because C = \frac{q}{v}} \right] \cr & \Rightarrow 3{C_1} = 5{C_2} \cr} $$

Graph (C) will be the right graph, the electric field inside the dielectrics will be less than the electric field outside the dielectrics. The electric field inside the dielectrics could not be zero.

As $${K_2} > {K_1},$$   the drop in electric field for $${K_2}$$ dielectric must be more than $${K_1}.$$

Since capacitance $$C = \frac{{{\varepsilon _0}A}}{d},$$   as $$d$$ decreases capacitance increases.

When $$S$$ is closed, there will be no shifting of negative charge from plate $$A$$ to $$B$$ as the charge $$ - q$$  is held by the charge $$ + q.$$  Neither there will be any shifting of charge from $$B$$ to $$A.$$