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.

Initial force in vacuum \(F_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . New force in medium \(F' = \frac{1}{4\pi\epsilon_0 K} \frac{(4q_1)(4q_2)}{(2r)^2} = \frac{16}{4K} \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}\) . Since \(F' = F_0\), we have \(frac{4}{K} = 1\), so \(K = 4\).

A completely symmetric distribution of \(n\) charges has a net field of zero at the center. Removing one charge is equivalent to superimposing a charge of \(-q\) at the empty corner on an otherwise full polygon, which produces a field of magnitude \(\frac{kq}{r^2}\).

The work done in rotating a dipole in a uniform electric field is given by \(W = PE(cos\theta_1 - cos\theta_2)\). Since it is rotated from \(\theta_1 = 0^\circ\) to \(\theta_2 = 90^\circ\), \(W = PE(cos 0^\circ - cos 90^\circ) = PE\).

A charge outside a closed surface contributes zero net flux (A-S). Gauss's law states that for an enclosed charge \(q\), the net flux is \(\frac{q}{\epsilon_0}\) (B-P). The electric field due to an infinite plane sheet is \(E = \frac{\sigma}{2\epsilon_0}\) (C-Q). For a charged conducting sphere, the external field is \(E = \frac{KQ}{r^2}\) (D-R).