What is dipole and electric field due to a dipole at a point on axial line and equatorial line.

Electric Dipole 

A pair of equal and opposite point charges that are separated by a small and finite distance is known as an electric dipole. 

Electric Dipole Moment

It is defined as the product of either charge and the distance between these two charges.

i.e.                                               

                                

 S.I. system of units, p is measured in coulomb-meter.

Electric Field Due To a Dipole

Electric Field Intensity due to an Electric Dipole at a Point on the Axial Line

Suppose that the electric dipole AB lies in the medium of the dielectric constant K . Let the dipole consists of two point charges -q and +q coulombs separated by a short distance of 2d meters. Let P be the observation point on the axial line such that its distance from the center of the point O of the electric dipole is r. Now Calculate the electric field strength at P.

                               

              

E1​=4πε0​1​(r−d)2q  

due to q at P {along the direction BP}

E2​=4πε0​1​(r+d)2q

due to -q at P {along the direction PA}

Intensities El and E2 are along the same line but in opposite direction. Since El > E2, the resultant intensity E at point  P will be equal to the difference and towards AP. Thus

 The resultant electric field at $P$ is

$E=E_1+\left(-E_2\right)$

$E=\left[\frac{1}{4\pi {\epsilon }_0}\frac{q}{{\left(r-d\right)}^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{{\left(r+d\right)}^2}\right]$ 

$E=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{{\left(r-d\right)}^2}-\frac{1}{{\left(r+d\right)}^2}\right]$ 

E=4πε0​q​
(r2−d2)24rd​]                            along BP

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

$\therefore E=\frac{q}{4\pi {\epsilon }_0}-\frac{4rd}{r^4}=\frac{q}{4\pi {\epsilon }_0}\frac{4d}{r^3}$

$E=\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}$ along with $BP$

             [$\because$ Electric dipole moment $p=q\times 2d$]

$E$ acts in the direction of dipole moment.

Electric Field Intensity due to an Electric Dipole at a Point on the equatorial plane 

 Suppose the observation point P lies on the equator of the dipole such that the distance from the midpoint O of the electric dipole is r. Let us assume again that the intermediate medium of the electric dipole and the observation point has a dielectric constant K. 

                                      

                           

                                          

$E_1=\frac{1}{4\pi {ϵ}_0}\genfrac{}{}{0ex}{}{q}{A{P}^2}$

$E_1=\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2}$ .........(x)         ($\because A{P}^2=O{A}^2+O{P}^2$)

Similarly,  $E_2$ at point P due to the south pole along PB is,

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{q}{P{B}^2}$

$E_2=\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2}$ .........(y)            ($\because P{B}^2=O{B}^2+O{P}^2$)

From equations (1) and (2), we get

$E_1=E_2=E(say)$

Now, the net electric field $E_P$ at P due to the dipole is

$E_P=E_1+E_2=2E$ ........................(z)

Resolving  $E_1$ and $E_2$ into horizontal and vertical components. The vertical components will be canceled and the horizontal will be added together. So, we can write 

$E_P=2E\cos \theta$  .............from(z)

$E_P=2\left(\frac{1}{4\pi {ϵ}_0}\frac{q}{(d^2+l^2{)}^2})(\frac{l}{\sqrt{(}{d}^2+l^2)}\right)$ ............($\because \cos \theta =\frac{l}{\sqrt{(}{d}^2+l^2)}$)

For a short dipole, $l<<d$, therefore

$B_P=\frac{1}{4\pi {ϵ}_0}\frac{2ql}{d^3}$

$B_P=\frac{1}{4\pi {ϵ}_0}\frac{Q}{d^3}$ ........................($\because Q=2ql$)

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