Q.Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately

The key idea is that at a large distance, any finite charge distribution looks like a point charge located at its centre. The potential therefore follows the inverse-distance law for a point charge.

1. For a collection of charges with total charge $Q_{\text{total}} = \sum q_i \neq 0$, the potential at a large distance $r$ is dominated by the monopole term.
2. The exact potential is $V = \frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{total}}}{r}$, plus smaller dipole and quadrupole corrections that fall off faster ($\propto 1/r^2$, $1/r^3$, etc.). …

The key idea is multipole expansion. When you are very far from a group of charges, the fine details of their arrangement blur out. The leading term in the potential is the monopole term, which depends only on the total charge $Q = \sum q_i$ and the distance $r$ from the centre of the distribution.

If $Q \neq 0$, then for large $r$ the potential is approximately

This is exactly the potential of a single point charge $Q$ placed at the centre of the distribution. For a point charge, the equipotential surfaces are spheres centred on the charge. Therefore, at great distances, the equipotentials are approximately spheres centred on the charge distribution.

1. Why the dipole and higher terms vanish at large $r$?

   The potential from a dipole falls off as $1/r^2$, from a quadrupole as $1/r^3$, and so on. When $r$ is huge, these terms become negligible compared to the $1/r$ monopole term. Only if the total charge were zero would the dipole term become the leading contribution, giving a different shape (like a peanut or a pair of lobes). But here the total sum is not zero, so the monopole dominates.

2. What does "approximately spherical" mean exactly?

   It means that if you draw an equipotential surface at a very large distance $R$, its shape deviates from a perfect sphere only by small ripples of order $1/R$ relative to the sphere's radius. As $R \to \infty$, the surface becomes exactly spherical. For any finite but large $R$, it is a slightly distorted sphere.

3. A concrete example …

Method: Far-Field (Multipole) Approximation for Equipotential Shape

Use this method whenever a question asks what equipotential surfaces look like "at a great distance" or "far from" a group of charges.

Steps

Step 1: Find the total charge of the distribution

Compute $Q_{\text{total}} = \sum q_i$. This single number decides which term in the multipole expansion survives at large distance.

Step 2: Identify the dominant (monopole) term when $Q_{\text{total}} \neq 0$

Far from any finite charge distribution, the fine details of how the charges are arranged blur out, and the potential is dominated by

exactly the potential of a single point charge $Q_{\text{total}}$ placed at the distribution's centre.

Step 3: Read off the equipotential shape from that dominant term …