$$\eqalign{ & {W_{{\text{el}}{\text{.}}}} = q\left( {{V_i} - {V_f}} \right) \cr & {\text{or}}\,\,6.4 \times {10^{ - 19}} = - 1.6 \times {10^{ - 19}}\left( {{V_A} - {V_B}} \right) \cr & {\text{or}}\,\,{V_A} - {V_B} = - 4\,V \cr & {\text{or}}\,\,{V_A} - {V_C} = - 4\,V\,\,\left( {\because {V_B} = {V_C}} \right) \cr & {\text{or}}\,\,{V_C} - {V_A} = 4\,V \cr} $$

The electron has negative charge. When an electron is brought towards another electron, then due to same negative charges repulsive force is produced between them. So, to bring them closer a work is done against this repulsive force. This work is stored in the form of electrostatic potential energy. Thus, electrostatic potential energy of system increases.
Alternative
Electrostatic potential energy of system of two electrons $$U = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( { - e} \right)\left( { - e} \right)}}{r}$$
$$ = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{r}$$
Thus, as $$r$$ decreases, potential energy $$U$$ increases.

Potential energy of the system
$$ = 4 \times \frac{K}{2}\left( {\frac{{{q^2}}}{L} + \frac{{{q^2}}}{L} + \frac{{{q^2}}}{{L\sqrt 2 }}} \right) = 5.4\frac{{K{q^2}}}{L}$$
= Final kinetic energy of the system
$$\eqalign{ & = 4 \times \frac{1}{2}m{v^2} = 2\,m{v^2} \cr & \therefore v = {\left[ {\frac{{K{q^2}}}{{mL}}\left( {2.7} \right)} \right]^{\frac{1}{2}}} \cr} $$

Surface of metallic cube is an equipotential surface. Therefore, electric field is normal to the surface of the cube.

Let charge on each sphere $$= q$$
when they are connected together their potential will be equal.
Now let charge on $$a = {q_1}$$  and $$b = 2q - {q_1}$$
$$\eqalign{ & \Rightarrow {V_a} = {V_b}\,\,{\text{or}}\,\,\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_1}}}{a} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{2q - {q_1}}}{b} \cr & \Rightarrow \frac{{{q_1}}}{{2q - {q_1}}} = \frac{a}{b} \cr & \frac{{{E_a}}}{{{E_b}}} = \frac{{\frac{{1}}{{4\pi {\varepsilon _0}}}\frac{{{q_1}}}{{{a^2}}}}}{{\frac{1}{{4\pi {\varepsilon _0}}}\frac{{{q_2}}}{{{b^2}}}}} = \left( {\frac{{{q_1}}}{{2q - {q_1}}}} \right)\frac{{{b^2}}}{{{a^2}}} \cr & = \frac{a}{b}.\frac{{{b^2}}}{{{a^2}}} = \frac{b}{a} = b:a \cr} $$

Volume of big drop $$ = 1000 \times {\text{volume}}\,{\text{of}}\,{\text{each}}\,{\text{small}}\,{\text{drop}}$$
$$\eqalign{ & \frac{4}{3}\pi {R^3} = 1000 \times \frac{4}{3}\pi {R^3} \Rightarrow R = 10r \cr & \because V = \frac{{kq}}{r}\,{\text{and}}\,V' = \frac{{kq}}{R} \times 1000 \cr} $$
Total charge on one small droplet is $$q$$ and on the big drop is $$1000\,q.$$
$$\eqalign{ & \Rightarrow \frac{{V'}}{V} = \frac{{1000r}}{R} = \frac{{1000}}{{10}} = 100 \cr & \therefore V' = 100\,V \cr} $$

As, shown in figure

If potential at centre is zero, then $${V_1} + {V_2} + {V_3} + {V_4} = 0$$
$$\eqalign{ & \Rightarrow - \frac{{kQ}}{r} - \frac{{kq}}{r} + \frac{{k2Q}}{r} + \frac{{k2q}}{r} = 0 \cr & \Rightarrow - Q - q + 2q + 2Q = 0 \cr & \therefore Q = - q \cr} $$

As work is done by the field, $$K.E.$$  of the body increases by
$$\eqalign{ & K.E. = W = q\left( {{V_A} - {V_B}} \right) \cr & = {10^{ - 8}}\left( {600 - 0} \right) \cr & = 6 \times {10^{ - 6}}\,J \cr} $$

Given, potential in a region, $$V = 6xy - y + 2yz.$$
Electric field in a region, $$E = - \frac{{\partial V}}{{\partial x}}\hat i - \frac{{\partial V}}{{\partial y}}\hat j - \frac{{\partial V}}{{\partial z}}\hat k$$
$$ \Rightarrow E = - 6y\hat i - \left( {6x - 1} \right)\hat j - 2y\hat k$$
At, $$\left( {1,1,0} \right),$$  electric field can be expressed,
$$\eqalign{ & E = - \left( {6 \times 1\hat i} \right) - \left( {6 \times 1 - 1} \right)\hat j - 2 \times 1.\hat k \cr & = - \left( {6\hat i + 5\hat j + 2\hat k} \right)N/C \cr} $$

$$\eqalign{ & eE = {m_e}{\omega ^2}r \cr & \Rightarrow \int E \,dr = \frac{{{m_e}{\omega ^2}}}{e}\int\limits_0^R {rdr} \cr & \Rightarrow V = \frac{{{m_e}{\omega ^2}{R^2}}}{{2e}} \cr} $$