Name: Continuous Charge Distribution
Uploaded: 2019-07-19
Duration: 5 min 39 s
Description: This video will help us understand the necessity of continuous charge distribution and learn about the concept of linear charge density. Learn more.

Question

Define the term electric dipole moment, and derive an expression for electric field intensity due to a dipole on its axial line.

Accepted Answer

Strength of an electric dipole is measured by its electric dipole moment, whose magnitude is equal to product of magnitude of either charge and separation between the two charges i.e., vector ( $p = q . 2 a$ ) and is directed from negative to positive charge, along the line joining the two charges. Its SI unit is $C m$ .

Consider an electric dipole with a dipole moment $\vec{P}$ as shown below. Magnitude of the dipole moment be

$P = 2 q a$

The two charges on the dipole are $2 a$ distance apart, as shown.

Let P be a point on the axial line of the dipole. Let the distance of point P from the midpoint of the dipole be r such that $r > > > a$ .

The electric field at point P will be due to the two charges on the given dipole.

The magnitude of electric field due to the charge $+ q$ will be $E_{+ q} = \frac{k q}{{(r - a)}^{2}}$

Since it is a positive charge, the direction of electric field $E_{+ q}$ will be along the direction of dipole moment (away from the charge)

The magnitude of electric field due to the charge $- q$ will be $E_{- q} = \frac{k q}{{(r + a)}^{2}}$

Since it is a negative charge, the direction of electric field $E_{- q}$ will be in the opposite direction of the dipole moment (towards the charge).

Therefore, the net electric field at point P will be $E = E_{+ q} - E_{- q}$

$E = \frac{k q}{{(r - a)}^{2}} - \frac{k q}{{(r + a)}^{2}} \Rightarrow E = k q (\frac{1}{{(r - a)}^{2}} - \frac{1}{{(r + a)}^{2}}) \Rightarrow E = k q (\frac{{(r + a)}^{2} - {(r - a)}^{2}}{{(r - a)}^{2} {(r + a)}^{2}}) \Rightarrow E = k q (\frac{r^{2} + 2 a r + a^{2} - (r^{2} - 2 a r + a^{2})}{((r - a) . {(r + a))}^{2}}) \Rightarrow E = k q (\frac{4 a r}{{(r^{2} - a^{2})}^{2}}) \Rightarrow E = 4 k q a (\frac{r}{(r^{4} - 2 a r + a^{4})})$

We can rewrite the above equation as,

$\Rightarrow E = 2 k (2 q a) (\frac{r}{(r^{4} - 2 a r + a^{4})})$

And $2 q a = P$

$\Rightarrow E = 2 k P (\frac{r}{(r^{4} - 2 a r + a^{4})}) . . . . . . . . . . . . (i)$

But, it is given that $r$ is very much greater than $a$ $(r > > > a)$ .

Therefore, in the expression $(r^{4} - 2 a r + a^{4})$ of equation (i), the term $- 2 a r + a^{4}$ is very much smaller than the term $r^{4}$ . When a very smaller number is added to a very large number, the sum will be almost equal to the larger number. Therefore, the smaller number does not contribute much to the sum and can be considered negligible.

Therefore, $(r^{4} - 2 a r + a^{4}) \approx r^{4}$ .
Now we can write the equation (i) as, $E = 2 k P (\frac{r}{r^{4}})$ .
The final value of net electric field is $E = 2 k P (\frac{1}{r^{3}})$ .

In the vector form, the net electric field will be written as $\vec{E} = 2 k P (\frac{1}{r^{3}}) . \hat{E}$ , where $\hat{E}$ is the unit vector of $\vec{E}$ .
The direction of the field is in the direction of the dipole moment. Therefore, the unit vector of the net electric field and the dipole moment vectors will be equal i.e. $\hat{E} = \hat{P}$ .

Now, the net electric is $\vec{E} = 2 k P (\frac{1}{r^{3}}) . \hat{P}$ .

The product of the magnitude of a vector and its unit vector gives us the vector. Therefore, $P . \hat{P} = \vec{P}$

This implies that $\vec{E} = 2 k (\frac{1}{r^{3}}) . \vec{P}$
Hence, the electric field due to a dipole, at a given point on its axial line is $\vec{E} = \frac{2 k}{r^{3}} . \vec{P}$ .

Define the term electric dipole moment, and derive an expression for electric field intensity due to a dipole on its axial line.

Important Questions on Electric Charges and Fields

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