Electric Dipole: Definition, Electric Field, Electric Potential & Torque on Dipole

The magnetic and electric dipole moments reveal structural information about the systems on the atomic and subatomic scale. On a large scale, in radio transmission and reception, dipole antennae are used. Do you know Electric Dipole occurs in nature in a variety of situations? For example, hydrogen fluoride molecule \(\left( {HF} \right).\) When a hydrogen atom combines with a fluorine atom, then a single electron of the previous is strongly attracted to the latter and spends most of its time near the fluorine atom. Although the molecule is electrically neutral, it has a dipole moment. Due to this dipole moment, the \(HF\) have some specific properties (Higher Boiling point, increased aqueous solubility).

What is Electric Dipole?

An electric dipole is a pair of equal and opposite point charges \(q\) and \(–q,\) separated by any fixed distance (let’s say \(2a\)). The line joining the two charges defines the length of the dipole, and the direction from \(–q\) to \(q\) is said to be the direction of the dipole according to sign convention. The mid-point of the line joining of \(–q\) and \(q\) is called the centre of the dipole.

Electric Dipole Moment \(\left( p \right):\) It is a vector quantity that measures the strength of an electric dipole by multiplying the charge \(\left( q \right)\) and separation between the charges \(\left( {2a} \right).\)
The magnitude of dipole moment is given as:
\(p = \left( {2a} \right)q\)

Direction: From negative charge \(\left( { – q} \right)\) to positive charge \(\left( { + q} \right)\)
SI unit: Its SI unit is \(\rm{C}\,\rm{m}\) (coulomb-meter).

Electric Field Due to a Dipole

As the total charge of the electric dipole is zero, but this does not mean that the field of the electric dipole is zero because the charge \(q\) and \(–q\) are apart by some distance hence if we add the electric fields due to them, it does not cancel out exactly. However, if the charges are separated by a distance much smaller than the distance of the field point \(\left( {2a < < r} \right),\) some approximations come in play and the fields due to \(q\) and \(–q\) nearly cancel out.

The electric field and the electric potential at any point in the vicinity of a dipole can be calculated just by adding the contributions due to each of the charges. From Coulomb’s law and the superposition principle, we can easily get the electric field of the pair of charges (\(–q\) and \(q\)) at any point in space. By using the parallelogram law of vectors, we can get the electric field at any general point \(P\) by adding the electric fields \({E_{ – q}}\) due to the charge \( – q\) and \({E_{ + q}}\) due to the charge \(q.\) The two cases are given below:

1. For points on the dipole axis

In the figure shown below, let us assume a point \(P\) at distance \(r\) from the centre of the dipole on the side of the charge \(q,\)

Then, we can write
\({E_{ – q}} = \, – \frac{q}{{4\pi \varepsilon {{\left( {r + a} \right)}^2}}}\widehat p\) ….(i)
where \(\widehat p\) is the unit vector along the dipole axis (from \(–q\) to \(q\)). Also, \({E_{ + q}} = \, – \frac{q}{{4\pi \varepsilon {{\left( {r – a} \right)}^2}}}\widehat p\) …(ii)
The total field at \(P\) by adding equations (i) and (ii) will be
\({E_{{\rm{axis}}}} = {E_{ + q}} + {E_{ – q}}\)
\({E_{{\rm{axis}}}} = \frac{q}{{4\pi \varepsilon }}\left( {\frac{1}{{{{\left( {r – a} \right)}^2}}} – \frac{1}{{{{\left( {r + a} \right)}^2}}}} \right)\widehat p\)
\({E_{{\rm{axis}}}} = \frac{q}{{4\pi \varepsilon }}\frac{{4ar}}{{{{\left( {{r^2} – {a^2}} \right)}^2}}}\widehat p\) …..(iii)
Now, for \(r > > a\)
\({E_{{\rm{axis}}}} = \frac{{4qa}}{{4\pi \varepsilon {r^3}}}\widehat p\)
Hence, we can also write this for \(\left( {r > > a} \right)\) and \(p = 2qa\) as:
\({E_{{\rm{axis}}}} = \frac{{2p}}{{4\pi \varepsilon {r^3}}}\) ….(iv)

2. For points on the equatorial plane of the dipole

The magnitudes of the electric fields due to the two charges \(+q\) and \(–q\) are equal, and it is given by
\({E_{ + q}} = \frac{q}{{4\pi \varepsilon }}\frac{1}{{{r^2} + {a^2}}}\) ….(v)
\({E_{ – q}} = \frac{q}{{4\pi \varepsilon }}\frac{1}{{{r^2} + {a^2}}}\) ….(vi)

The directions of \({E_{ + q}}\) and \({E_{ – q}}\) are shown in Fig. (b). Now after cancelling the components normal to the dipole axis and adding up the components along the dipole axis. The total electric field is opposite to \(\widehat p.\) We have
\(E = \, – \left( {{E_{ + q}} + {E_{ – q}}} \right)\,\cos \,\theta \widehat p\)
\(E = \, – \frac{{2qa}}{{4\pi \varepsilon {{\left( {{r^2} – {a^2}} \right)}^{\frac{3}{2}}}}}\widehat p\) …(vii)

At large distances \(\left( {r > > a} \right),\) this reduces to
\(E = \, – \frac{{2qa}}{{4\pi \varepsilon {r^3}}}\widehat p\) …(viii)
Hence, we can also write this for \(\left( {r > > a} \right)\) and \(p = 2qa\) as:
\(E = \, – \frac{p}{{4\pi \varepsilon {r^3}}}\) ….(ix)
where,
\(E\) is the electric field,
\(p\) is the electric dipole moment,
\(r\) is the distance of a point the point,
\(θ\) is the angle subtended by the dipole to the point.

Electrical Potential Due to Dipole

As the electric potential is a scalar quantity, so the electrical potential due to a dipole is the scalar sum of the potential of each charge separately.

1. At axial Position

Assume that a dipole is formed by two charges, \(–q\) and \(+q,\) separated by a distance of \(2a.\)

The electrical potential at the axial position at a point \(P\) at any distance \(r\) is given by
\({V_{{\rm{net}}}} = \frac{{kq}}{{{r_1}}} – \frac{{kq}}{{{r_2}}}\)
\({V_{{\rm{net}}}} = \left( {\frac{{kq}}{{\left( {r + a} \right)}} – \frac{{kq}}{{\left( {r – a} \right)}}} \right)\)
\({V_{{\rm{net}}}} = \, – \frac{{kq \times 2a}}{{{r^2} – {a^2}}}\)
For \(r > > a\)
\({V_{{\rm{net}}}} = \, – \frac{{kq}}{{{r^2}}}\) …(x)
where,
\(V\) is the electric potential,
\(r\) is the distance of a point of potential,
\(θ\) is the angle subtended by the dipole to the point.

2. For points on the equatorial plane

The electrical potential at the axial position at a point \(P\) at any distance \(r\) is given by
\({V_{{\rm{net}}}} = \,\frac{{kq}}{{{r_1}}} – \frac{{kq}}{{{r_2}}}\)
Since, \({r_1} = {r_2},\) we have
\({V_{{\rm{net}}}} = 0\)

Dipole in a Uniform Electric Field

As in the figure shown in Fig. (c), Let us consider a permanent dipole of dipole moment \(p\) in a uniform electric field \(E,\)), such that there is a force \(qE\) acting on \(q\) and a force \(–qE\) acting on \(–q.\) Then, the net force on the dipole is zero since \(E\) is uniform. As the charges are at a separate distance, so the forces will act be at different points, it results in a torque on the dipole.
The magnitude of torque will be given as,
\(\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \left( {\overrightarrow a \times \overrightarrow {qE} } \right) + \left( {\overrightarrow a \times \overrightarrow {qE} } \right)\)
\(\overrightarrow \tau = \overrightarrow {2qa} \times \overrightarrow E \)
\(\overrightarrow \tau = \overrightarrow p \times \overrightarrow E \)

And, its direction is normal to the plane of the paper, coming out of it, so we can also write this as:
\(\tau = pE\,\sin \,\theta \)
where,
\(\tau = \) The torque on the dipole,
\(E = \) The electric field.

Physical Significance of Dipoles

In most molecules, the centres of positive charges and negative charges lie in the same place. Therefore, their dipole moment is zero. \(C{O_2}\) and \(C{H_4}\) are of this type of molecule. However, when an electric field is applied to it then, they develop a dipole moment. But some molecules, have a permanent electric dipole moment, even in the absence of an electric field, Because the centres of negative charges and positive charges do not coincide in these molecules. Therefore, these molecules are called polar molecules. Water molecules, \({H_2}O,\) are an example of this type. It gives rise to interesting properties and important applications in the presence or absence of electric fields to the various materials.

Electric Dipole – Sample Problems

Q.1. Two charges \(±20\,\rm{μC}\) are placed at a distance of \(10\,\rm{mm}\). What is the electric field due to dipole at a point which is \(10\,\rm{cm}\) away from its centre \(O\) on the side of the positive charge?

Solution: Here, given
\(p = q\left( {2a} \right)\)
\( \Rightarrow p = \left( {20 \times {{10}^{ – 6}}} \right) \times \left( {10 \times {{10}^{ – 3}}} \right)\,{\rm{cm}}\)
\( \Rightarrow p = 2 \times {10^{ – 7}}\,{\rm{cm}}\)
\(r = 0.1\,{\rm{m}} = {10^{ – 1}}\,{\rm{m}}\)
Now,
Electric field due to an electric dipole on the axis at a distance \(r\) from the centre is given as:
\({E_{{\rm{axis}}}} = \frac{{2p}}{{4\pi \varepsilon {r^3}}}\) for \(\left( {r > > a} \right)\) and \(p = 2qa\)
\( \Rightarrow {E_{{\rm{axis}}}} = \frac{{2kp}}{{{r^2}}}\) where \(k = \frac{1}{{4\pi \varepsilon }}\)
\( \Rightarrow {E_{{\rm{axis}}}} = \frac{{2 \times 9 \times {{10}^9} \times 2 \times {{10}^{ – 7}}}}{{{{\left( {{{10}^{ – 1}}} \right)}^3}}}\)
\( \Rightarrow {E_{{\rm{axis}}}} = 3.6 \times {10^6}\,\rm{N}\,\rm{{C}^{ – 1}}.\)

Summary

An electric dipole is a pair of equal and opposite charges \(q\) and \(–q\) separated by some fixed distance. A dipole moment vector \(p\) has a magnitude of \(2qa,\) and it is in the direction of the dipole axis from \(–q\) to \(q.\)
Electric field due to an electric dipole on the axis at a distance \(r\) from the centre is given as:
\({E_{{\rm{axis}}}} = \frac{{2p}}{{4\pi \varepsilon {r^3}}}\) for \(\left( {r > > a} \right)\) and \(p = 2qa\)
Electric field due to an electric dipole in its equatorial plane at a distance \(r\) from the centre is given as:
\(E = \, – \frac{p}{{4\pi \varepsilon {r^3}}}\) for \(\left( {r > > a} \right)\) and \(p = 2qa\)
The magnitude of torque will be equals to the magnitude of each force multiplied by the arm of the couple, and its direction is normal to the plane of the paper, coming out of it as:
\(\tau = p \times E\)

FAQs on Electric Dipole

The answers to the most frequently asked questions on electric dipole are answered here:

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