If a conducting sphere of radius R is charged. Then the electric field at a distance r (r > R) from the centre of the sphere would be, (V = potential on the surface of the sphere):

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?