Here, $$V = 15 \times {10^6}\,V$$   dielectric strength
$$ = 5 \times {10^7}\,V\,{m^{ - 1}}$$
Maximum electric field,
$$E = 10\% $$  of dielectric strength
$$\eqalign{ & \therefore E = \frac{{10}}{{100}} \times 5 \times {10^7} = 5 \times {10^6}\,V\,{m^{ - 1}} \cr & {\text{As}}\,E = \frac{V}{r} \cr & \therefore r = \frac{V}{E} = \frac{{15 \times {{10}^6}}}{{5 \times {{10}^6}}} = 3\,m \cr} $$

As $$n$$ plates are joined, it means $$\left( {n - 1} \right)$$  capacitor joined in parallel.
∴ resultant capacitance $$ = \left( {n - 1} \right)C$$

The capacitance of a parallel plate capacitor with dielectric (oil) between its plates is
$$C = \frac{{K{\varepsilon _0}A}}{d}\,\,......\left( {\text{i}} \right)$$
where, $${{\varepsilon _0}} =$$  electric permittivity of free space
$$K =$$  dielectric constant of oil
$$A =$$  area of each plate of capacitor
$$d =$$  distance between two plates
When dielectric (oil) is removed, so capacitance of capacitor becomes, $${C_0} = \frac{{{\varepsilon _0}A}}{d}\,\,......\left( {{\text{ii}}} \right)$$
Comparing Eqs. (i) and (ii), we get
$$\eqalign{ & C = K{C_0} \cr & \Rightarrow {C_0} = \frac{C}{K} = \frac{C}{2}\,\,\left( {K = 2} \right) \cr} $$

From figure, $${Q_2} = \frac{2}{{2 + 1}}Q = \frac{2}{3}Q$$
$$\eqalign{ & Q = E\left( {\frac{{C \times 3}}{{C + 3}}} \right) \cr & \therefore {Q_2} = \frac{2}{3}\left( {\frac{{3CE}}{{C + 3}}} \right) = \frac{{2CE}}{{C + 3}} \cr} $$
Therefore graph D correctly dipicts.

Area of plate $$ = \pi {r^2} = \pi \times {\left( {8 \times {{10}^{ - 2}}} \right)^2}$$
$$ = 0.0201\,{m^2}$$
$${\text{and}}\,d = 1\,mm = 1 \times {10^{ - 3}}m$$
Capacity of capacitor
$$\eqalign{ & C = \frac{{{\varepsilon _0}{\varepsilon _r}A}}{d} = \frac{{8.85 \times {{10}^{ - 12}} \times 6 \times 0.0201}}{{1 \times {{10}^{ - 3}}}} \cr & = 1.068 \times {10^{ - 9}}F \cr} $$
Potential difference, $$V = 150\,volt$$
Energy stored,
$$\eqalign{ & U = \frac{1}{2}C{V^2} = \frac{1}{2} \times \left( {1.068 \times {{10}^{ - 9}}} \right) \times {\left( {150} \right)^2} \cr & = 1.2 \times {10^{ - 5}}J \cr} $$

Energy stored in a system of capacitors $$ = \sum {\frac{1}{2}C{V^2}} $$
Also, potential drop remains same in parallel across both capacitors. Initially stored energy $${U_1} = \frac{1}{2}C{V^2}$$

Finally, potential drop across each capacitor will be still $$V.$$
So, finally stored energy
$$\eqalign{ & {U_2} = \frac{1}{2}C{V^2} + \frac{1}{2}C{V^2} \cr & = \frac{1}{2}\left( {2C} \right){V^2} \cr & = 2\left( {\frac{1}{2}C{V^2}} \right) \cr & = 2{U_1} \cr} $$

Capacity of parallel plate capacitor
$$\eqalign{ & C = \frac{{{\varepsilon _r}{\varepsilon _0}A}}{d}\,\,\left( {{\text{For}}\,{\text{air}}\,{\varepsilon _r} = i} \right) \cr & {\text{So,}}\,\frac{{{\varepsilon _0}A}}{d} = 8 \times {10^{ - 12}} \cr & {\text{If}}\,d \to \frac{d}{2}\,{\text{and}}\,{\varepsilon _r} \to 6\,{\text{then}}\,{\text{new}}\,{\text{capacitance}} \cr & C' = 6 \times \frac{{{\varepsilon _0}A}}{{\frac{d}{2}}} = 12\frac{{{\varepsilon _0}A}}{d} = 12 \times 8\,pF = 96\,pF \cr} $$