Energy stored in a charged capacitor is in the form of electric field energy and it resides in the dielectric medium between the plates. This energy stored in the capacitor is given by $$U = \frac{1}{2}C{V^2}$$
If initial potential is $${V_1}$$ and final potential is $${V_2},$$  then increase in energy $$\left( {\Delta U} \right)$$
$$\eqalign{ & \Delta U = \frac{1}{2}C\left( {V_2^2 - V_1^2} \right) \cr & = \frac{1}{2} \times \left( {6 \times {{10}^{ - 6}}} \right) \times \left[ {{{\left( {20} \right)}^2} - {{\left( {10} \right)}^2}} \right] \cr & = \left( {3 \times {{10}^{ - 6}}} \right) \times 300 \cr & = 9 \times {10^{ - 4}}\,J \cr} $$

As potential at any point due to a point charge is given by,
$$\eqalign{ & V = \frac{{kQ}}{r}\,\,\left[ {k = \frac{1}{{4\pi {\varepsilon _0}}}} \right] \cr & V = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{r} \cr & E = \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{Q}{{{r^2}}} \cr & E = \frac{{4\pi {\varepsilon _0}{V^2}}}{Q} = 4\pi {\varepsilon _0} \times \frac{{{Q^2} \times {{10}^{22}}}}{Q} \cr & E = 4\pi {\varepsilon _0}Q \times {10^{22}}\,V/m \cr} $$

Potential at any point inside the sphere = potential at the surface of the sphere $$=10\,V.$$

Electrostatic potential energy of charge $$+q$$  placed at the centre of cube is $$U = 8 \times \frac{1}{{4\pi {\varepsilon _0}}} \times \frac{{q\left( { - q} \right)}}{{{\text{half - diagonal distance}}}}$$
$$\eqalign{ & = 8 \times \frac{1}{{4\pi {\varepsilon _0}}}\frac{{ - {q^2}}}{{b\frac{{\sqrt 3 }}{2}}}\left[ {_{{\text{where,}}\,b\, = \,{\text{side}}\,{\text{of}}\,{\text{cube}}}^{{\text{diagonal}}\,{\text{of}}\,{\text{cube}}\,\,{\text{ = }}\sqrt 3 b}} \right] \cr & = \frac{{ - 4{q^2}}}{{\sqrt 3 \pi {\varepsilon _0}b}} \cr} $$

Here, $$V\left( x \right) = \frac{{20}}{{{x^2} - 4}}volt$$
We know that $$E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}\left( {\frac{{20}}{{{x^2} - 4}}} \right)$$
$$\eqalign{ & {\text{or,}}\,\,E = + \frac{{40x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr & {\text{At}}\,\,x = 4\,\mu m, \cr & E = + \frac{{40 \times 4}}{{{{\left( {{4^2} - 4} \right)}^2}}} = + \frac{{160}}{{144}} = + \frac{{10}}{9}\,volt/\mu m. \cr} $$
Positive sign indicates that $${\vec E}$$ is in $$+ve$$  $$x$$-direction.

The electric field at a point is equal to negative of potential gradient at that point.
$$\eqalign{ & \vec E = - \frac{{\partial V}}{{\partial r}} = \left[ { - \frac{{\partial V}}{{\partial x}}\hat i - \frac{{\partial V}}{{\partial y}}\hat j - \frac{{\partial V}}{{\partial z}}\hat k} \right] \cr & = \left[ {\left( {2xy + {z^3}} \right)\hat i + \hat j\,{x^2} + \hat k\,3x{z^2}} \right] \cr} $$

$$\eqalign{ & {W_{PQ}} = q\left( {{V_Q} - {V_P}} \right) \cr & = \left( { - 100 \times 1.6 \times {{10}^{ - 19}}} \right)\left( { - 4 - 10} \right) \cr & = + 2.24 \times {10^{ - 16}}J \cr} $$

The potential at $$P$$ due to whole disc is
$$V = \frac{\sigma }{{2{ \in _0}}}\left[ {\sqrt {{R^2} + {r^2} - r} } \right].$$
Now potential due to quarter disc,
$$V = \frac{V}{4} = \frac{\sigma }{{8{ \in _0}}}\left[ {\sqrt {{R^2} + {r^2}} - r} \right].$$

$$\eqalign{ & U = 2k{q^2}\left[ { - \frac{1}{a} + \frac{1}{{2a}} - \frac{1}{{3a}} + \frac{1}{{4a}} + ......} \right] \cr & = - \frac{{2{q^2}}}{{4\pi {\varepsilon _0}a}}\left[ {1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ......} \right] \cr & = - \frac{{2{q^2}{{\log }_e}2}}{{4\pi {\varepsilon _0}a}} \cr} $$

Kinetic energy of bullet $$ = qV\,\,\left[ {_{V = {\text{ potential difference}}}^{q \,= \,{\text{charge on bullet}}}} \right]$$
From energy conservation, $$\frac{1}{2}m{v^2} = qV \Rightarrow V = \frac{{m{v^2}}}{{2q}}$$
Given, $$m = 2\,g = 2 \times {10^{ - 3}}\,kg,v = 10\,m/s,$$
$$q = 2\mu C = 2 \times {10^{ - 6}}\,C$$
Substituting the values in relation for $$V,$$ we obtain
$$\eqalign{ & V = \frac{{2 \times {{10}^{ - 3}} \times {{\left( {10} \right)}^2}}}{{2 \times 2 \times {{10}^{ - 6}}}} \cr & = 50 \times {10^3}\,V \cr & = 50\,kV \cr} $$