The equipotential surfaces due to an isolated point charge q < 0 are

The correct answer is Option (1) → concentric spherical shells and distance between the shells increases with the distance from point charge

The potential of a point charge depends only on the distance from it, \( V = \frac{q}{4\pi\varepsilon_0 r} \), so every point at the same r has the same potential. The equipotential surfaces are therefore concentric spheres centred on the charge.

To compare the spacing, take shells that differ by the same potential step \( \Delta V \). The field is the potential gradient, \( |E| = \Delta V/\Delta r \), and \( |E| = |q|/(4\pi\varepsilon_0 r^{2}) \) falls off as \( 1/r^{2} \).

    \( \displaystyle \Delta r = \frac{\Delta V}{|E|} = \frac{4\pi\varepsilon_0\,\Delta V}{|q|}\,r^{2} \)

So the gap between neighbouring shells grows as \( r^{2} \): the shells are crowded close to the charge, where the field is strong, and become farther apart with distance. The sign of q (here negative) only reverses the sign of the potential; the shape and the spacing are the same.

NCERT (2026-27):
Equipotential Surfaces – “An equipotential surface is a surface with a constant value of potential at all points on the surface. … This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge.”