Before we understand the characteristics of the electric potential of a dipole, let us quickly review our understanding of dipole and electric potential. An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point inside an electric field without producing acceleration. A dipole is a pair of opposite charges with equal magnitudes separated by a distance, d.
The electric potential due to a point charge q at a distance of r from that charge is given by,
\(V\) = \(\frac{1}{4πε_0}~\frac{q}{r}\)
The electric potential is a scalar field whose gradient becomes the electrostatic vector field. Since it is a scalar field, it is 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,
\(V\) = \(\sum\limits_{i}~V_i\) = \(V_{+}~+~V_{-}\)
\(V\) = \(\frac{1}{4πε_0}~\left(\frac{q}{r_{+}}~ +~\frac{-q}{r_{-}}\right)\)
\(V\) = \(\frac{q}{4πε_0}\left(\frac{1}{r_{+}}~ -~\frac{1}{r_{-}} \right)\)
\(V\) = \(\frac{q}{4πε_0}~\left(\frac{r_{-}~-~r_{+}}{r_{+}~ r_{-}}\right)\)
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,
\(r_{-}~-~r_{+}\) ≈ \(d cos~θ\)
Thus, we can write the potential as:
\(V\) = \(\frac{q}{4πε_0}~\left(\frac{d ~cos~θ}{r^2}\right)\)
We know that the magnitude of the electric dipole moment is:
\(|\overrightarrow{p}|\) = \(p\) = \(q~.~d\)
Thus, electric potential due to a dipole at a point far away from the dipole is given by,
\(V\) = \(\frac{1}{4πε_0}~ \frac{p~ cos~θ}{r^2}\)
Notice that the potential falls off by \(r^2\) while the electric field falls off by \(r^3\). Also note that when the angle is 90 degrees, the point \(P\) is equidistant to both charges and the electric potential is zero. When θ > 90⁰, the potential is negative because the point P is closer to the negative charge.
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