The electric potential due to a point charge q at a distance of r from that charge is given by,

\(V\)

The electric potential is a scalar field whose gradient becomes the electrostatic vector field. Being a scalar field, it is very easy to find the potential due to a system of charges. It is the summation of the electric potentials at a point due to individual charges.

Thus, we can write the net electric potential due to the individual potentials contributed by charges as

\(V_{net} = \frac{1}{4πε_0}~\sum\limits_{i}~\frac{q_i}{r_i}\)

Let us use this concept to find the electric field of a dipole. Let the distance between the point P and the positive and negative charges be r+ and r- respectively. Let the distance from the midpoint of the dipole be r. Let this vector subtend an angle θ to the dipole axis. Thus using the above theorem we have,

\(V\)

\(V\)

\(V\)

\(V\)

If the point P is sufficiently far from the dipole, then we can approximate

And by drawing a line perpendicular to r- from +q we can write,

\(r_{-}~-~r_{+}\)

Thus we can write the potential as:

\(V\)

We know that the magnitude of the electric dipole moment is:

\(|\overrightarrow{p}|\)

Thus, electric potential due to a dipole at a point far away from the dipole is given by,

\(V\)

Notice that the potential falls off by \(r^2\)

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