We know that potential energy of two charge system is given by
$$U = \frac{1}{{4\pi { \in _0}}}\frac{{{q_1}{q_2}}}{r}$$
According to question,
$$\eqalign{ & {U_A} = \frac{1}{{4\pi { \in _0}}}\frac{{\left( { + q} \right)\left( { - Q} \right)}}{a} = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Qq}}{a} \cr & {\text{and}}\,{U_B} = \frac{1}{{4\pi { \in _0}}}\frac{{\left( { + q} \right)\left( { - Q} \right)}}{a} = - \frac{1}{{4\pi {\varepsilon _0}}}\frac{{Qq}}{a} \cr & \Delta U = {U_B} - {U_A} = 0 \cr} $$
We know that for conservative force,
$$W = - \Delta U = 0$$

Potential should be same i.e. $${V_A} = {V_B}$$

$$V = \frac{{kq.2a\cos \theta }}{{{r^2}}} + \frac{{Kq}}{r} = \frac{{Kq}}{r}\left[ {\frac{{2a\cos \theta }}{r} + 1} \right]$$

Equipotential surfaces are normal to the electric field lines. The following figure shows the equipotential surfaces along with electric field lines for a system of two positive charges.

$$\eqalign{ & {q_r} + {q_R} = Q\,.......\left( 1 \right) \cr & \frac{{{q_r}}}{{{q_R}}} = \frac{{4\pi {r^2}\sigma }}{{4\pi {R^2}\sigma }} = \frac{{{r^2}}}{{{R^2}}} \cr & \Rightarrow \frac{{{q_r}}}{{{q_R}}} = \frac{{{r^2}}}{{{R^2}}}\,.......\left( 2 \right) \cr} $$
From (1) and (2)
$$\eqalign{ & {q_r} = \frac{{Q{r^2}}}{{{R^2} + {r^2}}}\,{\text{and}}\,{q_R} = \frac{{Q{R^2}}}{{{R^2} + {r^2}}} \cr & {\text{So,}}\,\,V = \frac{{{q_r}}}{{4\pi {\varepsilon _0}r}} + \frac{{{q_R}}}{{4\pi {\varepsilon _0}R}} \cr & \Rightarrow V = \frac{Q}{{4\pi {\varepsilon _0}}}\left( {\frac{{r + R}}{{{r^2} + {R^2}}}} \right) \cr} $$

In equilibrium, $$F = qE = \left( {n\,e} \right)\frac{V}{d} = mg$$
$$n = \frac{{mg\,d}}{{eV}} = \frac{{1.96 \times {{10}^{ - 15}} \times 9.8 \times 0.02}}{{1.6 \times {{10}^{ - 19}} \times 800}} = 3$$

Due to conducting sphere
At centre, electric field $$E = 0$$
And electric potential $$V = \frac{Q}{{4\pi { \in _0}R}}$$

When charge $${{q_3}}$$ is at $$C,$$ then its potential energy is $${U_C} = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.5}}} \right)$$
When charge $${{q_3}}$$ is at $$D,$$ then potential energy is $${U_D} = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_1}{q_3}}}{{0.4}} + \frac{{{q_2}{q_3}}}{{0.1}}} \right)$$
Hence, change in potential energy
$$\eqalign{ & \Delta U = {U_D} - {U_C} = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_2}{q_3}}}{{0.1}} - \frac{{{q_2}{q_3}}}{{0.5}}} \right) \cr & {\text{but}}\,\,\Delta U = \frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k \cr & \frac{{{q_3}}}{{4\pi {\varepsilon _0}}}k = \frac{1}{{4\pi {\varepsilon _0}}}\left( {\frac{{{q_2}{q_3}}}{{0.1}} - \frac{{{q_2}{q_3}}}{{0.5}}} \right) \cr & k = {q_2}\left( {10 - 2} \right) = 8{q_2} \cr} $$

$$\eqalign{ & E = 500\,V/m \cr & V = 3000\,V. \cr} $$
We know that electric field $$\left( E \right) = 500 = \frac{V}{d}$$
$${\text{or}}\,\,d = \frac{{3000}}{{500}} = 6\,m$$

We know that potential energy of discrete system of charges is given by
$$U = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{{q_1}{q_2}}}{{{r_{12}}}} + \frac{{{q_2}{q_3}}}{{{r_{23}}}} + \frac{{{q_3}{q_1}}}{{{r_{31}}}}} \right)$$
According to question,
$$\eqalign{ & {U_{{\text{initial}}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{{q_1}{q_2}}}{{0.3}} + \frac{{{q_2}{q_3}}}{{0.5}} + \frac{{{q_3}{q_1}}}{{0.4}}} \right) \cr & {U_{{\text{final}}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{{q_1}{q_2}}}{{0.3}} + \frac{{{q_2}{q_3}}}{{0.1}} + \frac{{{q_3}{q_1}}}{{0.4}}} \right) \cr & {U_{{\text{final}}}} - {U_{{\text{initial}}}} = \frac{1}{{4\pi { \in _0}}}\left( {\frac{{{q_1}{q_2}}}{{0.1}} - \frac{{{q_2}{q_3}}}{{0.5}}} \right) \cr & = \frac{1}{{4\pi { \in _0}}}\left[ {10{q_2}{q_3} - 2{q_2}{q_3}} \right] = \frac{{{q_3}}}{{4\pi { \in _0}}}\left( {8{q_2}} \right) \cr} $$