Electric Potential Of A Dipole And System Of Charges

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 an 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,
\(V\) = \(\frac{1}{4πε_0}~\frac{q}{r}\)

Where, \(ε_0\) is the permittivity of free space.

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

\(V_{net} = \sum\limits_{i}~V_{i}\)

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

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,

\(V\) = \(\sum\limits_{i}~V_i\) = \(V_{+}~+~V_{-}\)

\(V\) = \(\frac{1}{4πε_0}~\left(\frac{q}{r_{+}}~ +~\frac{-q}{r_{-}}\right)\)

Electric Potential

\(V\) = \(\frac{q}{4πε_0}\left(\frac{1}{r_{+}}~ -~\frac{1}{r_{-}} \right)\)

\(V\) = \(\frac{q}{4πε_0}~\left(\frac{r_{-}~-~r_{+}}{r_{+}~ r_{-}}\right)\)

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

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

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.

 


Practise This Question

Two charges of equal magnitude 'q'  but of opposite sign separated by a distance 2a constitute an electric dipole of dipole moment p. If P is a point at a distance r from the centre of the dipole and the line joining the centre of the dipole to this point makes an angle θ with the axis of the dipole, then the potential at the point P  is given by  (r2a) (Where p=2qa )