Question

Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Answer 1

Hint: To derive the expression for electric field due to an electric dipole, consider AB is an electric dipole of two point charges – q and + q separated by small distance 2d and then consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

Formula used: $E={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{x^2}}$

Complete step-by-step answer:

We have been asked to derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Now, let AB be an electric dipole of two-point charges -q and + q separated by a small distance 2d.

Also, consider that P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

Now, for better understanding refer the figure below:

The electric field at any point due to a charge q at a distance x is given by-

$E={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{x^2}}$ 

Now, the electric field at the point P due to + q charge placed at B is given by-

$E_1={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{{(r-d)}^2}}$  (along BP)

Here, (r – d) is the distance of point P from charge +q.

Also, the electric field at point due to – q charge placed at A-

$E_2={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{{(r+d)}^2}}$  (along PA)

Therefore, the magnitude of the resultant electric field (E) acts in the direction of the vector with a greater magnitude.

The resultant electric field at P is-

$E=E_1+(-E_2)$ 

Putting the values; we get-

$E=\left[{\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{{(r-d)}^2}}-{\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{q}{{(r+d)}^2}}\right]$  (along BP)

Simplifying further we get- 

$E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.\left[{\displaystyle \frac{1}{{(r-d)}^2}}-{\displaystyle \frac{1}{{(r+d)}^2}}\right]$

$E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.\left[{\displaystyle \frac{{(r+d)}^2-{(r-d)}^2}{{(r-d)}^2{(r+d)}^2}}\right]$ 

Now, we can write $$(r-d{)}^2(r+d{)}^2=(r^2-d^2{)}^2$$

So, we get-

$E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.\left[{\displaystyle \frac{{(r)}^2+{(d)}^2+2rd-{(r)}^2-{(d)}^2+2rd}{{(r-d)}^2{(r+d)}^2}}\right]$

$E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.\left[{\displaystyle \frac{4rd}{{(r^2-d^2)}^2}}\right]$ 

Now, if the point P is far away from the dipole, then $$d\ll r$$

So, the electric field will be-

$E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.\left[{\displaystyle \frac{4rd}{{(r^2)}^2}}\right]={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{4rd}{r^4}}={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{4d}{r^3}}$  

along BP

Also, electric dipole moment  $p=q\times 2d$  , so the expression will now become-

 $E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{2(2d)}{r^3}}={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{2p}{r^3}}$ 

E acts in the direction of the dipole moment.

Therefore, the expression for the electric field intensity due to an electric dipole at a point on its axial line is $E={\displaystyle \frac{q}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{2(2d)}{r^3}}={\displaystyle \frac{1}{4\pi {\in }_{\circ }}}.{\displaystyle \frac{2p}{r^3}}$ 

Note: 

Whenever it is required to derive an expression, then always first draw a rough sketch depicting a dipole and a point on the axial line. As mentioned in the figure, first we found out the electric field at the point P due to + q charge placed at B and then due to charge – q placed at A. Then, the resultant electric field at P is found out. After making necessary assumptions, we got the expression for electric field intensity.