Electric Potential due to an Electric Dipole:
The electric potential due to an electric dipole can be measured at different points:
1.) The electric potential on the axis of the electric dipole
2.) The electric potential on the equatorial line of the electric dipole
3.) The electric potential at any point of the electric dipole
1. The electric potential on the axis of the 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 a point $P$ be on the axis of an electric dipole and placed 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 potential due to the dipole's charges.
So Electric potential at point $P$ due $+q$ charge of electric dipole→
$ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{r-\frac{l}{2}}$
The electric potential at point $P$ due $-q$ charge of electric dipole→
$ V_{-q}=-\frac{1}{4\pi \epsilon_{0}} \frac{q}{r+\frac{l}{2}}$
Electric potential is a scalar quantity. Hence, the resultant potential $V$ at the point $P$ will be the algebraic sum of the potential $V_{+q}$ and $V_{-q}$. i.e. →
$ V=V_{+q}+V_{-q}$
Now substitute the value of $V_{+q}$ and $V_{-q}$ in the above equation →
$ V= \frac{1}{4\pi \epsilon_{0}} \frac{q}{r-\frac{l}{2}} -\frac{1}{4\pi \epsilon_{0}} \frac{q}{r+\frac{l}{2}}$
$ V= \frac{1}{4\pi \epsilon_{0}} \left[ \frac{q}{r-\frac{l}{2}} - \frac{q}{r+\frac{l}{2}} \right]$
$ V= \frac{q}{4\pi \epsilon_{0}} \left[ \frac{1}{r-\frac{l}{2}} - \frac{1}{r+\frac{l}{2}} \right]$
$ V= \frac{q}{4\pi \epsilon_{0}} \left[ \frac{ \left( r+\frac{l}{2} \right)-\left (r-\frac{l}{2} \right)}{r^{2}-\frac{l^{2}}{4}} \right]$
$ V= \frac{1}{4\pi \epsilon_{0}} \left[ \frac{ql}{r^{2}-\frac{l^{2}}{4}} \right]$
$ V= \frac{1}{4\pi \epsilon_{0}} \left[ \frac{p}{r^{2}-\frac{l^{2}}{4}} \right] \qquad \left( \because p=ql\right)$
If $r$ is much larger than $l$. So $\frac{l^{2}}{4}$ can be neglected in comparison to $r^{2}$. Therefore, the electric potential at the point $P$ due to the electric dipole is →
$ V= \frac{1}{4\pi \epsilon_{0}} \left[ \frac{p}{r^{2}} \right] $
2. The electric potential on the equatorial line of the 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 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 potential due to the dipole's charges.
So Electric potential at point $P$ due $+q$ charge of electric dipole→
$ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{BP}$
$ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{\sqrt{r^{2}+\frac{l^{2}}{4}}}$
The electric potential at point $P$ due $-q$ charge of electric dipole→
$ V_{-q}=-\frac{1}{4\pi \epsilon_{0}} \frac{q}{AP}$
$ V_{-q}=-\frac{1}{4\pi \epsilon_{0}} \frac{q}{\sqrt{r^{2}+\frac{l^{2}}{4}}}$
$\therefore$ The resultant potential at point $P$ is
$ V=V_{+q}+V_{-q}$
$ V=\frac{1}{4\pi \epsilon_{0}} \frac{q}{\sqrt{r^{2}+\frac{l^{2}}{4}}}-\frac{1}{4\pi \epsilon_{0}} \frac{q}{\sqrt{r^{2}+\frac{l^{2}}{4}}} $
$V=0 $
Thus, the electric potential is zero on the equatorial line of a dipole (but the intensity is not zero). So, no work is done in moving a charge along this line.
3. The electric potential at any point of the 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 its distance is $r$ from the center point $O$ of the electric dipole AB.
Let the distance of point $P$ from the point $A$ and Point $B$ of the dipole is $PB=r_{1}$ and $PA=r_{2}$ respectively.
So, the electric potential at point $P$ due to the $+q$ charge of the electric dipole is →
$ V_{+q}=\frac{1}{4\pi \epsilon_{0}} \frac{q}{r_{1}}$
$ V_{-q}=-\frac{1}{4\pi \epsilon_{0}} \frac{q}{r_{2}}$
The resultant potential at point $P$ is the algebraic sum of the potential due to charges $+q$ and $-q$ of the dipole. That is
$ V=V_{+q}+V_{-q}$
$ V=\frac{1}{4\pi \epsilon_{0}} \frac{q}{r_{1}}-\frac{1}{4\pi \epsilon_{0}} \frac{q}{r_{2}}$
$ V=\frac{1}{4\pi \epsilon_{0}} \left(\frac{q}{r_{1}}-\frac{q}{r_{2}} \right) \qquad(1)$
Now simplify the above equation by applying the Geometry from the above figure. i.e. From the figure, Acute angle $\angle POB$, we can write as,
$ r^{2}_{1}=r^{2}+\frac{l^{2}}{4}-rlcos\theta \qquad(2)$
Similarly for displacement $\overrightarrow{r_{2}}$
$ r^{2}_{2}=r^{2}+\frac{l^{2}}{4}-rlcos \left(\pi - \theta \right)$
$ r^{2}_{2}=r^{2}+\frac{l^{2}}{4}+rlcos \theta \qquad(3)$
The equation $(2)$ may be expressed as →
$ r^{2}_{1}=r^{2} \left[1+ \frac{l^{2}}{4r^{2}}-\frac{l}{r}cos\theta \right] $
Taking distance $r$ much greater than the length of dipole (i.e. r>>l), so we may retain only the first order term in $\frac{l}{r}$,
$ \therefore r^{2}_{1}=r^{2} \left[1- \frac{l}{r}cos\theta \right]$
$ r_{1}=r \left[1- \frac{l}{r}cos\theta \right]^{\frac{1}{2}}$
$ \frac {1}{r_{1}}=\frac{1}{r} \left[1- \frac{l}{r}cos\theta \right]^{-\frac{1}{2}}$
Now, applying the binomial theorem to the above equation. So we get
$ \frac {1}{r_{1}}=\frac{1}{r} \left[1+ \frac{l}{2r}cos\theta \right]$
Similarly,
$ \frac {1}{r_{2}}=\frac{1}{r} \left[1- \frac{l}{2r}cos\theta \right]$
Substituting these values in equation $(1)$, we get
$ V=\frac{1}{4\pi\epsilon_{0}} \left[ \frac{q}{r} \left(1+ \frac{l}{2r}cos\theta \right)-\frac{q}{r} \left(1- \frac{l}{2r}cos\theta \right)
\right]$
$ V=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r} \left[ \left(1+ \frac{l}{2r}cos\theta \right)- \left(1- \frac{l}{2r}cos\theta \right) \right]$
$ V=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r} \left[ \left(1+ \frac{l}{2r}cos\theta \right)- \left(1- \frac{l}{2r}cos\theta \right) \right]$
$ V=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r} \left[ \frac{l cos\theta}{r}\right]$
$ V=\frac{1}{4\pi\epsilon_{0}} \left[ \frac{ql cos\theta}{r^{2}}\right]$
But $q\times l=p$ (dipole moment)
$ V=\frac{1}{4\pi\epsilon_{0}} \left[ \frac{p cos\theta}{r^{2}}\right]$
The vector form of the above equation can be written as →
$ V=\frac{1}{4\pi\epsilon_{0}} \left[ \frac{\overrightarrow{p} \cdot \overrightarrow{r} }{r^{3}}\right]$
The above two equations hold only under the approximation that the distance of the observation point $P$ is much greater than the size of the dipole.
Special Case:
- At axial points $\theta=0^{\circ}$,
then $cos\theta= cos 0^{\circ}=1$,
Therefore, $ V=\frac{1}{4\pi\epsilon_{0}} \frac{p}{r^{2}}$
- At equatorial points $\theta=90^{\circ}$,
then $cos\theta= cos 90^{\circ}=0$,
Therefore,$ \quad V=0$
Now comparing this result with the potential due to a point-charge, we see that:
- In a fixed direction, that is, fixed $\theta$, $V\propto \frac{1}{r^{2}}$. here rather than $V \propto \frac{1}{r}$;
- Even for a fixed distance $r$, there is now a dependence on direction, that is, on $\theta$.
