What is the use of dipole moments?

strength of the dipole in the field

What will be the potential due to the dipole at a point P?

difference of potential at charge

product of potentials at charge

Electric Field Of A Dipole - Definition, Formula, Examples and More

What Is the Electric Field of a Dipole?

A dipole is a separation of opposite electrical charges and it is quantified by an electric dipole moment. The electric dipole moment associated with two equal charges of opposite polarity separated by a distance, d is defined as the vector quantity having a magnitude equal to the product of the charge and the distance between the charges and having a direction from the negative to the positive charge along the line between the charges.

It is a useful concept in dielectrics and other applications in solid and liquid materials. These applications involve the energy of a dipole and the electric field of a dipole.

How to Calculate Electric Field of a Dipole?

Consider an electric dipole with charges +q and -q separated by a distance d.

Electric Field of a Dipole

We shall for the sake of simplicity only calculate the fields along symmetry axes, i.e. a point P along the perpendicular bisector of the dipole at a distance r from the mid-point of the dipole and a point Q along the axis of the dipole at a distance r from the mid-point of the dipole.

Along the Perpendicular Bisector (Point P)

The electric fields due to the positive and negative charges (Coulomb’s law):

\begin{array}{l} E_{+} = \frac{1}{4 π ε_{0}} \frac{q}{r_{+}^{2}} \end{array}

\begin{array}{l} = \frac{1}{4 π ϵ_{0}} \frac{q}{(\sqrt{r^{2} + (\frac{d}{2})^{2}})^{2}} \end{array}

\begin{array}{l} = \frac{1}{4 π ϵ_{0}} (\frac{q}{r^{2} + (\frac{d}{2})^{2}}) \end{array}

Similarly,

\begin{array}{l} E_{-} = \frac{1}{4 π ε_{0}} \frac{q}{r_{-}^{2}} = \frac{1}{4 π ε_{0}} \frac{q}{r^{2} + (\frac{d}{2})^{2}} \end{array}

The vertical components of the electric field cancel out as P is equidistant from both charges.

\begin{array}{l} ⟹ E = E_{+} c o s θ + E_{-} c o s θ \end{array}

\begin{array}{l} ⟹ E = \frac{1}{4 π ε_{0}} \frac{q}{r^{2} + (\frac{d}{2})^{2}} c o s θ + \frac{1}{4 π ε_{0}} \frac{q}{r^{2} + (\frac{d}{2})^{2}} c o s θ \end{array}

\begin{array}{l} ⟹ E = \frac{1}{4 π ε_{0}} \frac{2 q}{r^{2} + (\frac{d}{2})^{2}} c o s θ \end{array}

Now,

\begin{array}{l} c o s θ = \frac{\frac{d}{2}}{r_{+}} = \frac{\frac{d}{2}}{r_{-}} = \frac{\frac{d}{2}}{\sqrt r^{2} + (\frac{d}{2})^{2}} \end{array}

Substituting this value we get,

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{2 q}{r^{2} + (\frac{d}{2})^{2}} \frac{\frac{d}{2}}{\sqrt r^{2} + (\frac{d}{2})^{2}} = \frac{1}{4 π ε_{0}} \frac{q d}{(r^{2} + (\frac{d}{2})^{2})^{\frac{3}{2}}} \end{array}

Dipole moment

\begin{array}{l} p = q \times d \end{array}

When r >> d, we can neglect the d/2 term. Thus, we have,

\begin{array}{l} E = \frac{1}{4 π ε_{0}} {\frac{p}{r^{2}}}^{\frac{3}{2}} \end{array}

\begin{array}{l} ⟹ E = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}} \end{array}

The dipole moment direction is defined as pointing towards the positive charge. Thus, the direction of the electric field is opposite to the dipole moment:

\begin{array}{l} \vec{E} = - \frac{1}{4 π ε_{0}} \frac{\vec{p}}{r^{3}} \end{array}

Along Axis of Dipole (Point Q)

The electric fields due to the positive and negative charges are:

\begin{array}{l} E_{+} = \frac{1}{4 π ϵ_{0}} \frac{q}{r_{+}^{2}} = \frac{1}{4 π ϵ_{0}} \frac{q}{(r - \frac{d}{2})^{2}} E_{-} = \frac{1}{4 π ϵ_{0}} \frac{q}{r_{-}^{2}} = \frac{1}{4 π ϵ_{0}} \frac{q}{r + {\frac{d}{2}}^{2}} \end{array}

Since the electric fields are along the same line but in opposing directions,

\begin{array}{l} E = E_{+} - E_{-} \end{array}

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{q}{(r - \frac{d}{2})^{2}} - \frac{1}{4 π ε_{0}} \frac{q}{(r + \frac{d}{2})^{2}} \end{array}

\begin{array}{l} E = \frac{q}{4 π ε_{0}} [\frac{1}{(r - \frac{d}{2})^{2}} - \frac{1}{(r + \frac{d}{2})^{2}}] \end{array}

\begin{array}{l} E = \frac{q}{4 π ε_{0}} [\frac{(r + \frac{d}{2})^{2} - (r - \frac{d}{2})^{2}}{(r^{2} - (\frac{d}{2})^{2})^{2}}] \end{array}

\begin{array}{l} E = \frac{q}{4 π ε_{0}} [\frac{4 r \frac{d}{2}}{(r^{2} - (\frac{d}{2})^{2})^{2}}] \end{array}

\begin{array}{l} E = \frac{1}{4 π ε_{0}} [\frac{2 r q d}{(r^{2} - (\frac{d}{2})^{2})^{2}}] \end{array}

\begin{array}{l} E = \frac{1}{4 π ε_{0}} [\frac{2 r p}{(r^{2} - (\frac{d}{2})^{2})^{2}}] \end{array}

Factoring r⁴ from denominator:

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{1}{r^{4}} [\frac{2 p r}{(1 - (\frac{d}{2 r})^{2})^{2}}] \end{array}

Now if r >> d, we can neglect the (d/2r)² term becomes very much smaller than 1. Thus, we can neglect this term. The equation becomes:

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{1}{r^{4}} [\frac{2 p r}{1^{2}}] \end{array}

\begin{array}{l} E = \frac{1}{4 π ε_{0}} \frac{2 p}{r^{3}} \end{array}

Since in this case the electric field is along the dipole moment, E₊ > E_–,

\begin{array}{l} \vec{E} = \frac{1}{4 π ε_{0}} \frac{2 \vec{p}}{r^{3}} \end{array}

Notice that in both cases the electric field tapers quickly as the inverse of the cube of the distance. Compared to a point charge which only decreases as the inverse of the square of the distance, the dipoles field decreases much faster because it contains both a positive and negative charge. If they were brought to the same point their electric fields would cancel out completely but since they have a small distance separating them, they have a feeble electric field.

Watch the video and learn about electric field due to dipole

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Frequently Asked Questions – FAQs

What is an electric dipole?

An electric dipole is defined as a couple of opposite charges q and –q separated by a distance d. By default, the direction of electric dipole in space is always from negative charge -q to positive charge q. The midpoint q and –q is called the centre of the dipole. The simplest example of an electric dipole is a pair of electric charges of two opposite signs and equal magnitude separated by distance.

What is the SI unit of the dipole moment?

The SI unit of dipole moment is Coulomb.meter

Give an example of an electric dipole.

A pair of electric charges of two opposite signs and equal magnitude separated by a distance.

How does an atom behave as a magnetic dipole?

The electrons in an atom revolve around the nucleus in a closed orbit. The orbit around the nucleus is equivalent to a current loop as the electrons are charged particles. The electrons revolve in anticlockwise while the current revolves in the clockwise direction. This movement of electrons creates a south pole and north pole resulting in the atom’s behaviour as a magnetic dipole.

What is electric charge?

Electric Charge is the property of subatomic particles that causes it to experience a force when placed in an electric and magnetic field.

To learn more related concepts like semiconductors, diode, etc., download BYJU’S – The Learning App.

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