The potential energy of a charged capacitor before removing the dielectric slat is $$U = \frac{{{Q^2}}}{{2C}}$$
The potential energy of the capacitor when the dielectric slat is first removed and the reinserted in the gap between the plates is $$U = \frac{{{Q^2}}}{{2C}}$$
There is no change in potential energy, therefore work done is zero.

For an isolated sphere, the capacitance is given by
$$C = 4\pi { \in _0}r = \frac{1}{{9 \times {{10}^9}}} \times 1 = 1.1 \times {10^{ - 10}}F$$

$$\eqalign{ & {V_0} = \frac{q}{{{C_0}}} \cr & V = \frac{q}{C} \Rightarrow \frac{V}{{{V_0}}} = \frac{{{C_0}}}{C} \cr & \Rightarrow \frac{{{C_0}}}{C} = \frac{{500}}{{75}} = \frac{{20}}{3} \cr} $$
By definition, $$C = k{C_0}$$
$$ \Rightarrow k = \frac{{20}}{3}$$

Energy required to charge the capacitor is $$W = U = QV$$
$$\eqalign{ & \Rightarrow U = C{V^2} = \frac{{{\varepsilon _0}A}}{d}.{V^2} = \frac{{{\varepsilon _0}Ad}}{{{d^2}}} \cr & {V^2} = {\varepsilon _0}{E^2}Ad\,\,\left[ {\because E = \frac{V}{d}} \right]. \cr} $$

If we thraw the charged particle just right of the center of the tunnel, the particle will cross the tunnel. Hence, applying conservation of momentum between start point and center of tunnel we get
$$\eqalign{ & \Delta K + \Delta U = 0 \cr & {\text{or}}\,\left( {0 + \frac{1}{2}m{v^2}} \right) + q\left( {{V_f} - {V_i}} \right) = 0 \cr & {\text{or}}\,{V_f} = \frac{{{V_s}}}{2}\left( {3 - \frac{{{r^2}}}{{{R^2}}}} \right) = \frac{{\rho {R^2}}}{{6{\varepsilon _0}}}\left( {3 - \frac{{{r^2}}}{{{R^2}}}} \right) \cr} $$
Hence $$r = \frac{R}{2}$$
$$\eqalign{ & {V_f} = \frac{{\rho {R^2}}}{{6{\varepsilon _0}}}\left( {3 - \frac{{{R^2}}}{{4{R^2}}}} \right) = \frac{{11\rho {R^2}}}{{24{\varepsilon _0}}};{V_i} = \frac{{\rho {R^2}}}{{3{\varepsilon _0}}} \cr & \frac{1}{2}m{v^2} = 1\left[ {\frac{{11\rho {R^2}}}{{24{\varepsilon _0}}} - \frac{{\rho {R^2}}}{{3{\varepsilon _0}}}} \right] \cr & = \frac{{\rho {R^2}}}{{{\varepsilon _0}}}\left[ {\frac{{11}}{{24}} - \frac{1}{3}} \right] \cr & = \frac{{\rho {R^2}}}{{8{\varepsilon _0}}}\,\,{\text{or}}\,\,V = {\left( {\frac{{\rho {R^2}}}{{4m{\varepsilon _0}}}} \right)^{\frac{1}{2}}} \cr} $$
Hence velocity should be slightly greater than $$V.$$

In series arrangement charge on each plate of all the capacitors have same magnitude. The potential difference is distributed inversely in the ratio of capacitors, i.e.
$$V = {V_1} + {V_2} + {V_3}\,\,\left[ {\because {V_1} = {V_2} = {V_3} = V} \right]$$
Here, $$V = 3V$$
The equivalent capacitance $${C_s}$$ is given by

$$\eqalign{ & \frac{1}{{{C_s}}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} + \frac{1}{{{C_3}}}\,\,\left[ {\because {C_1} = {C_2} = {C_3} = C} \right] \cr & {C_s} = \frac{C}{3} \cr} $$

Initially we know that
$$\eqalign{ & \Delta U = \frac{1}{2}\frac{{{C_1}{C_2}}}{{{C_1} + {C_2}}}{\left( {{V_1} - {V_2}} \right)^2} \cr & \Delta U = \frac{1}{2} \times \frac{{C \times C}}{{2C}}{\left( {{V_1} - {V_2}} \right)^2} \cr & \Delta U = \frac{C}{4}{\left( {{V_1} - {V_2}} \right)^2} \cr} $$

Common potential $$V' = \frac{{{C_1}V + {C_2} \times 0}}{{{C_1} + {C_2}}}$$
$$ = \frac{{{C_1}}}{{{C_1} + {C_2}}}V$$

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by
$$U = \frac{1}{2}\frac{{{Q^2}}}{C} = \frac{1}{2} \times \frac{{{{\left( {8 \times {{10}^{ - 18}}} \right)}^2}}}{{100 \times {{10}^{ - 6}}}} = 32 \times {10^{ - 32}}J$$

The equivalent capacitance of $$n$$ identical capacitors of capacitance $$C$$ is equal to $$nC.$$ Energy stored in this capacitor
$$E = \frac{1}{2}\left( {nC} \right){V^2} = \frac{1}{2}nC{V^2}$$