Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector P of magnitude, $4\times10^{-6}$ C m, is $\pm 9\times10^3$ V. (Take $1/4\pi\varepsilon_0 = 9\times10^9$ SI units) Reason (R): $V = \pm \frac{2P}{4\pi\varepsilon_0 r^2}$, where r is the distance of any axial point situated at 2 m from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

The variation of electrostatic potential with radial distance $r$ from the centre of a positively charged metallic thin shell of radius $R$ is given by the graph:

Six charges +q, -q, +q, -q, +q and -q are fixed at the corners of a hexagon of side d as shown in the figure. The work done in bringing a charge q₀ to the centre of the hexagon from infinity is: (ε₀ - permittivity of free space)

A, B and C are three points in a uniform electric field. The electric potential is:

The electric potential at a point in free space due to a charge $Q$ coulomb is $Q \times 10^{11}$ V. The electric field at that point is:

Four point charges -Q, -q, 2q and 2Q are placed, one at each corner of the square. The relation between Q and q for which the potential at the centre of the square is zero, is

If potential in a region is expressed as $V(x,y,z) = 6xy - y + 2yz$, the electric field at point $(1, 1, 0)$ is:

Two spheres of radius $a$ and $b$ respectively are charged and joined by a wire. The ratio of the electric field at the surface of the spheres is:

Two metal spheres, one of radius $R$ and the other of radius $2R$ respectively have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?