If the plates of a parallel plate capacitor connected to a battery are moved closer to each other, then: (A) The charge stored in it increases. (B) The energy stored in it decreases. (C) Its capacitance increases. (D) The ratio of charge to its potential remains the same. (E) The product of charge and voltage increases.

Choose the most appropriate 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?