Electric Dipole:
When two equal (i.e., equal in magnitude) and opposite-charge particles are placed at a very small distance apart, the system is known as an electric dipole.
Electric Dipole Moment:
The product of magnitude of one point charged particle and the distance between the charges is called the 'electric dipole moment'. It is vector quantity and the direction of electric dipole moment is along the axis of the dipole pointing from negative charge to positive charge.
Let us consider the two charged particles, which have equal magnitude $+q$ coulomb and $-q$ coulomb, placed at a very small distance of $l$ in a dipole, so the electric dipole moment is →
$\overrightarrow{p}=q\times \overrightarrow{l}$
Unit: $C-m$ Or $Ampere-metre-sec$
Dimension: $[ALT]$
Electric field intensity due to an Electric Dipole:
The electric field intensity due to an electric dipole can be measured at three different points:
- 1.) Electric field intensity at any point on the axis of an electric dipole
- 2.) Electric field intensity at any point on the equatorial line of an electric dipole
- 3.) Electric field intensity at any point on an electric dipole
1. Electric field intensity at any point on the axis of an electric dipole:
Let us consider an electric dipole $AB$ made up of two charges of $-q$ and $+q$ coulomb placed in a vacuum or air at a very small distance of $l$. Let take a point $P$ on the axis of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put a test charged particle $q_{0}$ at point $P$ for the measurement of the electric field intensity due to the dipole's charge.
So, the electric field intensity (magnitude only) at point $P$ due to a $+q$ charge of an electric dipole
$ E_{+q}=\frac{1}{4\pi\epsilon}\frac{q}{(r-\frac{l}{2})^{2}} \qquad(1)$
Electric field intensity (magnitude only) at point $P$ due $-q$ charge of electric dipole
$ E_{-q}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{(r+\frac{l}{2})^{2}} \qquad(2)$
The net electric field at point $P$ due to an electric dipole
$ E=E_{+q}-E_{-q}\qquad(3)$
Subtitute the value of $E_{+q}$ and $E_{-q}$ in equation $(3)$, Then the above equation $(3)$ can also be written as follows
$E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{1}{(r-\frac{l}{2})^{2}}-\frac{1}{(r+\frac{l}{2})^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r+\frac{l}{2})^{2}-(r-\frac{l}{2})^{2}}{(r+\frac{l}{2})^{2}(r-\frac{l}{2})^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{(r^{2}+\left(\frac{l}{2}\right)^{2}+2\frac{l}{2}r-r^{2}-\left(\frac{l}{2}\right)^{2}+2\frac{l}{2}r)}{(r+\frac{l}{2})^{2}(r-\frac{l}{2})^{2}} \right ]$
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{2rl}{(r^{2}-\frac{l^{2}}{4})^{2}} \right ]\qquad(4)$
Here $l \lt r $ so $\frac{l^{2}}{4} \lt \lt r^{2}$ therefore neglect the term $\frac{l^{2}}{4}$ in above equation $(4)$ so we can write above equation
$ E=\frac{q}{4\pi\epsilon _{0}}\left [ \frac{2rl}{(r^{2})^{2}} \right ]$
$E=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2p}{r^{3}} \right ]\qquad (5)$
This is the equation of electric field intensity at a point on the axis of an electric dipole.
The vector form of the above equation $(5)$ is
$\overrightarrow{E}=\frac{1}{4\pi\epsilon _{0}}\left [ \frac{2\overrightarrow{p}}{r^{3}} \right ]$
2. Electric field intensity at any point on the equatorial line of an electric dipole:
Let us consider an electric dipole $AB$ made up of two charges of $+q$ and $-q$ coulomb placed in vacuum or air at a very small distance $l$. Let a point $P$ be on the equatorial line of an electric dipole and place it at a distance $r$ from the center point $O$ of the electric dipole. Now put the test charged particle $q_{0}$ at point $P$ for the measurement of the electric field intensity due to the dipole's charge.
So, the electric field intensity (magnitude only) at point $P$ due to a $+q$ charge of an electric dipole
$E_{+q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ] \qquad(1)$
Electric field intensity (magnitude only) at point $P$ due to $-q$ charge of electric dipole
$E_{-q}=\frac{1}{4\pi\epsilon_{0}}\left [\frac{q}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ] \qquad(2)$
The net electric field at point $P$ due to an electric dipole
$E=E_{+q}\:cos\theta + E_{-q}\: cos\theta \qquad(3)$
Put the value of $E_{+q}$ and $E_{-q}$ in the above equation $(3)$, So equation $(3)$ can also be written as follows
$E=2\left [ \frac{q}{4\pi\epsilon_{0} }\frac{1}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}} \right ]cos\theta \qquad (4)$
From figure, In $\Delta \: POB$,
$cos\:\theta=\frac{l}{\sqrt{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}}}$
Put the value of $cos\theta$ in equation $(4)$, so equation $(4)$ can also be written as follows
$E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times l}{\left\{r^{2}+\left(\frac{l}{2}\right)^{2}\right\}^{3/2}} \right ] \qquad (5)$
Here $l \lt r$ so $\frac{l^{2}}{4} \lt \lt r^{2}$ so neglect the term $\frac{l^{2}}{4}$ in above equation $(5)$ so we can write above equation
$ E=\frac{1}{4\pi\epsilon_{0}}\left [ \frac{q\times l}{r^{3}} \right ] \qquad (6)$
$E=\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{3}} \qquad (7)$
This is the equation of electric field intensity at a point on the equatorial line of an electric dipole.
The vector form of the above equation $(7)$ is
$\overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}} \frac{\overrightarrow{p}}{r^{3}}$
3. Electric field intensity at any point of an electric dipole:
Let us consider an electric dipole $AB$ of length $l$ consisting of the charge $+q$ and $-q$. Let's take a point $P$ in general, and denote its position vector by $\overrightarrow{r}$ from the center point $O$ of the electric dipole AB to point $P$.
The electric dipole moment is a vector quantity that has a direction from $-q$ charge to $+q$ charge. So the electric dipole moment's direction is resolved into two components, one component $ pcosθ$ is along the vector position $\overrightarrow{r}$, and the other component $psinθ$ is normal to vector position $\overrightarrow{r}$. So
Electric field intensity due to dipole moment of component
$p\:cos\theta$ {Electric field along the Axial Line }
$ \overrightarrow{E_{\parallel }}=\frac{1}{4\pi\epsilon_{0}} \frac{2pcos\theta}{r^{3}} \qquad(1)$
Electric field intensity due to dipole moment of component
$p\:sin\theta$ {Electric field along the Equatorial Line}
$ \overrightarrow{E_{\perp}}=\frac{1}{4\pi\epsilon_{0}} \frac{psin\theta}{r^{3}} \qquad(2)$
The resultant electric field vector $E$ at point $P$
$ \overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2}+2 E_{\perp}E_{\parallel}cos90^{\circ}}$
From figure, The angle between $E_{\perp}$ and $E_{\parallel}$ is $90^{\circ}$. So
$\overrightarrow{E}=\sqrt{E_{\perp}^{2}+E_{\parallel}^{2} }\qquad (3)$
Now substitute the value of equation $(1)$ and equation $(2)$ in equation $(3)$. Then
$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(sin^{2}\theta+4cos^{2}\theta)}$
$ \overrightarrow{E}=\frac{1}{4\pi\epsilon_{0}}\frac{p}{r^{3}}\sqrt{(1+3cos^{2}\theta)}$
This is the equation of electric field intensity at any point due to an electric dipole.
The direction of the resultant electric field intensity vector $\overrightarrow{E}$ from the axial line is
$tan\alpha =\frac{\overrightarrow{E}_{\perp }}{\overrightarrow{E_{\parallel }}} \qquad(4)$
Put the value of $\overrightarrow{E_{\perp}}$ and $\overrightarrow{E_{\parallel}}$ in equation (4). we get
$tan\alpha =\frac{sin\theta}{2cos\theta}$
$tan\alpha =\frac{1}{2}tan\theta$
Here $\alpha$ is the angle between the resultant electric field intensity $\overrightarrow{E}$ and the axial line.
