What is Electric Dipole? Find Comprehensive Class 12 Notes

The Physics Class 12 book uses this definition. 

“An electric dipole is a pair of equal and opposite point charges q and –q, separated by a distance 2a.” (p. 23, Electric Charges and Fields, NCERT).

Three aspects of an electric dipole that are easy to remember:

• One positive charge is denoted by +q.
• Negative charge is denoted by -q.
• Line between the two charges with a distance between charges 2a.

Properties of an electric dipole to understand the definition more comprehensively:

• The meaning of equal charges is that their magnitudes are the same. 
• This arrangement of opposite charges means they have opposite polarities, like +q and - q. If one charge is positive, the other has to be negative. In this system, you cannot have the same charge signs. 
• An electric dipole’s total charge remains zero as the system is neutral. As the charges are separated, the electric fields do not really cancel each other out. The field is not entirely zero.

To really learn how the electric dipoles interact with their environment when the distances are larger, there are a few insights to look out for. 

Electric fields do not totally cancel each other out even when the dipole has two equal and opposite charges. 

So, if the electric fields from the two charges don't completely cancel out at a distance, how do we precisely describe the electric field at a specific point in space relative to the dipole itself?

Dipole Length vs Observation Distance

Here, we need to know the difference between dipole Length (2a) and observation distance (r)

You know the dipole length is 2a (sometimes denoted d). This separation exists between the two charges +q and -q. This is the size of the dipole. 

Now, for calculations at the human scale, we need to use another observation distance from where, you, the observer, is at from the dipole. It has the symbol r. 

Another way to see it is that the observation distance defines the position where the measurement of the electric field is taken, and that is relative to the dipole size and structure. 

The relationship, then, is that the observation distance r is much greater than 2a - the dipole length where we show as r>>2a. It’s too far away and always will be larger than the dipole. Naturally, even the midpoint distance of the dipole 2a/2 is a, which denotes partial cancellation of the electric field strength, and r is still greater than a, ie., r>>a. 

The Decay of the Electric Field Depends on Observation Distance

The strength of the field and distance relationship is different for a point charge and a dipole. 

1. If you recall the formula from Coulomb’s Law, for a single point charge, you know it follows the inverse square law in relation to the distance. That is always 1/r^2. Same for field lines that radiate outwards at this distance, and the field strength is in this inverse-squared proportion of 1/r^2.
2. But when we look at the electric field strength of a dipole in relation to the observation distance, it is 1/r^3. We should instead say that the field strength decreases by 1/r^3. 

So when it comes to macroscopic distances, where r>>a, the general magnitude of the electric field |E| due to a separation of the dipole is dependent on the 1/r^3 relationship.
We will show how it holds true for calculations of electric field due to a dipole. 

 

In this section on defining an electric dipole mathematically, we’re going to look at what effect the electric field (E) has in the presence of a dipole.
The simplest approaches here would be to look at two geometries: along the axis and equatorial plane of the dipole to learn how the electric field is affected. 

Practice and master these with Class 12 physics Chapter 1 NCERT Solutions. 

Field on Axial Line of Dipole

Here, the consideration is a short dipole where the observation distance (r) is much greater than distance separating the charges, ie., 2a. And we know r>>2a. 

The electric field (E) at some point on the axis or the line that connects the two charges, directed along the axis. Here, we are considering the direction from negative to positive (a concept of direction and sign convention we will cover while explaining electric dipole moment). The direction is then, -q to +q. 

We consider a point P at a distance of r from the dipole centre, where we split 2a into a. And p is the unit vector along the axis, moving from -q to +q.
Now, the field due to the negative charge (-q) gives us a distance of (r+a). The field of -q is E_-q

So in the vector form of electric field, E, we get the equation

E_-q = q/4πε_0(r+a)^2 x p

 

Think for the field of +q now, what would be the distance?

E_+q = q/4πε_0(r-a)^2 x p

It’s (r-a), as the charge +q is at an r distance from the midpoint of the axis and directed away from the positive charge and it’s along p.
So total electric field,  

E_total = E_-q + E_+q

This is the vector sum of the individual fields.

 

E_total = q/4πε_0 [ 1/ (r+a)^2 - 1/(r-a)^2 ] p

Now we simplify the common denominator to get

E_total = q/4πε_0 [ (r+a)^2 - (r-a)^2 / (r^2 - a^2)^2 ] p

 

And expanding the numerator, we get 4ar

For now, we can say that the magnitude of the electric field

|E| ≅ (q/4πε_0) x 4ar/r^4

Then to simplify we get, |E| ≅ (q/4πrε_0) x 4a/r^3

When r is much greater than a.

Field Points on the Equatorial Plane

On the equatorial plane, the magnitude of the electric field exerted by a dipole, is shown by,

|E| ≅ (q/4πrε_0) x 2a/r^3

The equatorial plane lies perpendicular to the axis of the centre of the dipole. So the magnitudes of the electric field for both +q and -q are equal. And when we add the total electric field E_total

E_total = q/4πε_0 [ 1/ (r+a)^2 - 1/(r+a)^2 ] p

Following the same simplification of the denominator and expanding the numerators, like above, we get the same inverse-cubic relationship. (Try doing that following the same above methods yourself!)

What to Remember for Exams?

• The electric field along the axis is exactly twice the field strength on the equatorial plane. 
• Unlike the single point charge whose electric field is dependent on distance in an inverse-square relationship (1/r^2), the electric dipole’s field strength for large distances (r>>2a) follows an inverse cubic (1/r^3) relationship. Meaning the field strength falls off much more due to a dipole than a single point charge. This is how we should understand about the term, electric field decay. 

The dipole moment of an electric dipole provides a measure of polarity, which means it shows strength and direction. Since it is a vector quantity, it indicates the direction of polarity. It is calculated as the product of the magnitude of the charge and the distance between these two charges (separation distance). 

So, mathematically,

$\vec{p}=q\cdot 2\vec{a}$

where:

$\vec{p}$ = dipole moment

$q$ = magnitude of each charge

$2a$ = distance between charges

Unit of Dipole Moment

The dipole moment SI unit is Coulomb-metre or Cm. 
You can learn more about Coulomb is the derived SI Unit of electric charge and learn the definition of Coulomb in detail.   

What is the Direction of Electric Dipole Moment?

We fixed the direction of the dipole moment according to a convention agreed upon (by the scientific world). The scientific community of physicists has fixed the direction which negative charge to positive charge for convenience. (For chemistry students, the direction goes from positive to negative that helps understand how electrons are distributed at the molecular level.)  

If you draw a line from the negative charge towards the positive charge, colinear to the axis of the dipole, this will represent the direction of the dipole moment.

Calculating Field Strength for Electric Dipole Moment

Now, if we use the calculations for the electric field due to a dipole axially and on the equatorial plane, we get the following relationships using the formula of the electric dipole moment. 

For axial, we replace 4aq with dipole moment equation: p = 2 x (q x 2a), while keeping the unit vector of position p  as it is

E = (q/4πrε_0) x 4a/r^3 p = (q/4πrε_0) x  2p/r^3 p

Same, if we do for the equatorial plane, we get,

E = (q/4πrε_0) x p/r^3 p

Also, remember the reference point is the centre of the dipole. The system's centre presents neutrality and it is generally what's used for calculations in this chapter. 

Assume there is an electric dipole with two charges: +q and -q. These charges are separated by a distance d.

Dividing this distance d in a way that the O is midpoint and the distance of the charge from it is r. 

Then the formula for electric potential

The electric potential due to a dipole: $\begin{array}{l}V=\frac{1}{4\pi ϵ}\frac{pcos\theta }{r^2}\end{array}$

In case, θ = 90°

Electric Potential (V) = 0

When, θ = 0°

$\begin{array}{l}\text{Electric potential}=V=\frac{1}{4\pi ϵ}\frac{p}{r^2}\end{array}$

Until now, we have been focusing on the dipole’s own or internal electric field. It’s the system we have analysed using Coulomb’s Law. We checked how the arrangement defines its surroundings, whether it’s axial or on the equatorial plane, when a test charge is placed nearby.
From here on, you should learn briefly about the interaction of the dipole when there is an external field. This becomes necessary to learn when a dipole encounters another charge configuration.

You might want to revisit a few topics of Class 11 Physics to understand the different surrounding scenarios.

Forces and Torques

Here, you have to think about what would happen when a dipole is placed in a uniform or non-uniform external field?

Uniform External Field

If it is a uniform electric field, the charges will cancel out by zeroing the total translational force as they are equal and opposite.  But there will be some torque that will make the dipole align with the field.
Mathematically, we just have get to using the torque formula, using the cross product of vector concept. The resulting torque vector should be perpendicular to the plane, which has a dipole moment p (the position vector) and the electric field E (applied force) - both vectors. 

If torque formula is τ = r x F

Equate that to

τ = p x E

If you are only looking at calculating the magnitude of the torque, use this formula: 

τ = pEsinθ

Non-Uniform External Field

In a non-uniform external field, the dipole experiences a torque. But there are two conditions of orientation here, which make the total torque to be zero. 

Condition 1: The dipole moment, suppose, is parallelly aligned to the non-uniform electric field, then the angle θ = 0 degree. So torque is zero.

Condition 2: If it’s antiparallel, the dipole moment creates an angle θ = 180 degrees.
Then comes the imbalance of forces on the opposite charges at different points resulting in different magnitudes and directions. As forces cannot cancel out, the total translational force of the dipole in a non-uniform external field becomes non-zero. 

You can solve these two practice questions on an electric dipole. 

1. A dipole with two charges of equal and opposite charges with magnitude $q=2\mu \mathrm{C}$ separated by a distance $2a=0.01\mathrm{m}$ in a uniform field $E=1000{\mathrm{N}\mathrm{C}}^{-1}$ . The angle between E and the axis of the dipole $\theta ={60}^{\circ }$ . Find the torque.

Ans. 

$\begin{array}{c}p=q\cdot 2a=2\times {10}^{-6}\times 0.01=2\times {10}^{-5}\mathrm{C}\mathrm{m}\\ \tau =pE\mathrm{s}\mathrm{i}\mathrm{n}{60}^{\circ }=\left(2\times {10}^{-5}\right)\times 1000\times \frac{\sqrt[]{3}}{2}\approx 0.0173\mathrm{N}\mathrm{m}.\end{array}$

2. What will be the electric field at a point 0.1 m away from a dipole that has a dipole moment $p=1\times {10}^{-6}\mathrm{C}\mathrm{m}$ on its axial line?

Ans. 

$E=\frac{1}{4\pi {ϵ}_0}\frac{2p}{r^3}=9\times {10}^9{\mathrm{N}\mathrm{m}}^2{\mathrm{C}}^{-2}\times \frac{2\times {10}^{-6}}{(0.1{)}^3}=1800{\mathrm{N}\mathrm{C}}^{-1}.$