Since potential is constant on an equipotential surface, the work done in moving a charge is zero \(W = q \Delta V = 0\). This is because the electric field is always perpendicular to the displacement along the surface \(\vec{E} \cdot d\vec{r} = 0\).

Initially, \( F = k \frac{q^2}{r^2}\). When 50% of the charge of one sphere is transferred to the other, the charges become \(0.5q\) and

\(1.5q\). The new force is \(F' = k \frac{(0.5q)(1.5q)}{r^2} = 0.75 F = \frac{3}{4}F\).

Since charges repel, they reside entirely on the outer surface of metallic conductors. Thus, the charge distribution is identical for hollow and solid metallic spheres of the same radius.

In a uniform electric field, the net force on a dipole is always zero. The potential energy is \(U = -pE \cos(0^\circ) = -pE\), which is the minimum potential energy.

Inside a hollow metal sphere, the electric field is zero. This implies the electric potential is constant throughout the interior and equal to the surface potential of 10 V.

When two conductors are connected by a conducting path (like a wire), charge flows between them until they reach electrostatic equilibrium, which means they acquire the same electrical potential.

When connected, their electric potentials become equal, i.e., \(V_1 = V_2\). Since \(V = \frac{\sigma R}{\varepsilon_0}\), we get \(\sigma_1 R_1 = \sigma_2 R_2\), which gives \(\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}\).

By volume conservation, \(R = N^{1/3} r = 27^{1/3} r = 3r\). The total charge of the combined drop is \(Q = 27q\). The potential of the bigger drop is \(V' = \frac{kQ}{R} = \frac{k(27q)}{3r} = 9 V = 9 \times 220 = 1980\text{ V}\).