The Electric Dipole

The Electric Dipole

The combination of two equal and opposite charges separated by a small distance form the electric dipole. Consider two charges +q and -q separated by a small distance \(|\vec{d}|\) then electric dipole moment of this combination is defined as a vector \(\vec{p}=q\vec{d}\) is the vector joining the negative charge to positive charge. The line along the direction of dipole moment is called axis of the dipole.

Electric Potential and Field due to a Dipole (Axis of dipole passes through the origin)

Electric potential

Suppose the negative charge -q is placed at a point A and the positive charge +q is placed at a point B on the figure. The separation of charge is d. The middle point of AB is origin O. The potential is to be evaluated at a point P where OP=r and <POB=\(\theta \) and r>>d.

Let AA' be the perpendicular from A to PO and BB' be the perpendicular from B to PO.

As d<<r, AP\(\approx \)A'P

\begin{align*}=OP+OA' \end{align*}\begin{align*}=r+OA.cos\theta [\because cos\theta =\frac{A'O}{AO}] \end{align*}\begin{align*} =r+\frac{d}{2}cos\theta\end{align*}\begin{align*}Similarly,\space BP \approx B'P \end{align*}\begin{align*}=OP-OB'=r-\frac{d}{2}cos\theta \end{align*}

Now the potential at the point due to negative charge -q is

\begin{align*}V_1=\frac{1}{4\pi \epsilon_0}\frac{-q}{r+\frac{d}{2}cos\theta} \rightarrow 1\end{align*}

Similarly, the potential due to +q is\begin{align*}V_2=\frac{1}{4\pi \epsilon_0}\frac{q}{r-\frac{d}{2}cos\theta}\rightarrow 2 \end{align*}

The net potential at point P due to dipole is \begin{align*}V=V_1+V_2 \end{align*}\begin{align*}=\frac{1}{4\pi \epsilon_0}\left [\frac{q}{r-\frac{d}{2}cos\theta}\frac{q}{r+\frac{d}{2}cos\theta} \right ] \end{align*}\begin{align*}=\frac{1}{4\pi \epsilon_0}\frac{dcos\theta}{\left ( r^2-\frac{d^2}{4}cos^2\theta \right )} \end{align*}\begin{align*}=\frac{qdcos\theta}{4\pi \epsilon_0 r^2}{\because r>>d} \end{align*}\begin{align*}=\frac{pcos\theta}{4\pi \epsilon_0 r^2}\rightarrow 3 \end{align*}

Here the potential at a distance r from a point charge q is given by

\begin{align*}V=\frac{q}{4\pi \epsilon_0 r} \end{align*}

which is inversely proportional to r and is independent of direction. But the potential due to dipole given by 3 is inversely proportional to \(r^2 \) and depends upon the direction as shown by the term \(cos\theta \).

In general any charge distribution that produces electric dipole potential is given by \begin{align*}V=\frac{pcos\theta}{4\pi \epsilon_0 r^2} \end{align*}

where p is the electric dipole moment and the direction from which angle \(\theta \) is measured to get the above equation is called the direction of dipole moment.

Electric Field

As in the given figure, two opposite charges +q and -q at B and B' respectively from an electric dipole and the point P is consideration point at a distance r from the centre of axis of dipole where we have to calculate electric field \(\vec{E}\). \(PP_1\) is a small displacement in the direction OP and \(PP_2\) is a small displacement perpendicular to OP.

\begin{align*}Here,\space OP=r,<POB=\theta \end{align*}\begin{align*}PP_1=dr,PP_2=rd\theta \end{align*}

Hence \(PP_1\) is in radial direction (dr) and \(PP_2\) is in transverse direction.

In going from P to \(P_1\), the angle \(\theta \) does not change and the distance OP changes from r+dr. In going from P to \(P_2\), the angle \(\theta \) changes to \(\theta \)+d\(\theta \), while the distance r remains almost constant. \begin{align*}Thus\space PP_2 =rd\theta \end{align*}From equation \begin{align*}E=-\bigtriangledown V \end{align*}

The component of the electric field at P in the radial direction PP1 is \begin{align*}E_{r}=-\frac{\partial V}{\partial r}\rightarrow 1 \end{align*}

The symbol \(\sigma \) specifies that \(\theta \) should be treated as constant while differentiating w.r.to r.

Similarly, the component of the electric field at P in the transverse direction \( PP_2\) is \begin{align*}E_{\theta}=-\frac{\partial V}{r\partial \theta}\rightarrow 2 \end{align*}

we know\begin{align*}E_{r}=-\frac{\partial V}{\partial r}=-\frac{1}{4\pi \epsilon_0}\frac{\partial }{\partial r}\left (\frac{pcos\theta}{r^2} \right ) \end{align*}\begin{align*}=-\frac{1}{4\pi \epsilon_0}.pcos\theta \frac{\partial r^{-2}}{\partial r}\end{align*}\begin{align*}=\frac{1}{4\pi \epsilon_0}\frac{2pcos\theta }{r^3} \rightarrow 3\end{align*}\begin{align*}And\space E_{\theta}=-\frac{\partial V}{r\partial \theta}\end{align*}\begin{align*}-\frac{1}{4\pi \epsilon_0} \frac{\partial }{r\partial \theta}\left (\frac{pcos\theta}{r^2} \right ) \end{align*}\begin{align*}=\frac{psin\theta}{4\pi \epsilon_0 r^3}\rightarrow 4 \end{align*}

The resultant electric field at P is given by \begin{align*}E=\sqrt{E_{r}^2+ E_{\theta}^2} \end{align*}\begin{align*}=\sqrt{\left ( \frac{1}{4\pi \epsilon_0}\frac{2pcos\theta }{r^3}\right )^2+\left ( \frac{psin\theta}{4\pi \epsilon_0 r^3}\right )^2} \end{align*}\begin{align*} =\frac{p}{4\pi \epsilon_0 r^3}\sqrt{4cos^2\theta+sin^2\theta}\end{align*}\begin{align*} =\frac{p}{4\pi \epsilon_0 r^3}\sqrt{3cos^2\theta +1}\rightarrow 5 \end{align*}

If the resultant field makes an angle \(\alpha \) with the radial direction OP, we have \begin{align*}tan\alpha=\frac{E_{\theta}}{E_{r}} =\frac12 tan\theta\end{align*}\begin{align*}\therefore \alpha =tan^{-1}[\frac12 tan\theta] \rightarrow 6\end{align*}

Electric Field and Potential (Origin is not on the axis of dipole)

Two equal and opposite charges separated by a finite distance form an electric dipole. Let charges -q and +q be located at positions \(\vec{r'}\) and \(\vec{r'}+\vec{l}\) with respect to origin O as shown in figure.

Now electric field at P whose position vector is \(\vec{r}\) relative to origin O is given by

\begin{align*}E(\vec{r})=\frac{q}{4 \pi \epsilon_0}\frac{(\vec{r}-\vec{r'}-\vec{l})}{|\vec{r}-\vec{r'}-\vec{l}|^3}-\frac{q}{4 \pi \epsilon_0}\frac{(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3} \rightarrow 1\end{align*}

As the dipole separation \(\vec{l} \) is a small as compared with \(\vec{r}-\vec{r'} \).

Hence from expansion,

\begin{align*}|\vec{r}-\vec{r'}-\vec{l}|^{-3}=[(\vec{r}-\vec{r'}-\vec{l})^2 ]^{-3/2} \end{align*} \begin{align*}=[(\vec{r}-\vec{r'})^2-2(\vec{r}-\vec{r'}).\vec{l}+\vec{l}^2]^{-3/2} \end{align*}\begin{align*}=\left [|\vec{r}-\vec{r'}|^2\left \{ 1-\frac{2(\vec{r}-\vec{r'}).\vec{l}}{|\vec{r}-\vec{r'}|^2} +\frac{\vec{l}^2}{|\vec{r}-\vec{r'}|^2} \right \}\right ]^{-3/2} \end{align*}\begin{align*}=|\vec{r}-\vec{r'}|^{-3}\left [1-\frac{-3}{2}\left \{ \frac{2(\vec{r}-\vec{r'}).\vec{l}}{|\vec{r}-\vec{r'}|^2} -\frac{\vec{l}^2}{|\vec{r}-\vec{r'}|^2} \right \}+... \right ] \end{align*}

As \(|\vec{l}|<<|\vec{r}-\vec{r'}|\), we can neglect \(l^2\) and higher power of it. So,\begin{align*}= |\vec{r}-\vec{r'}|^{-3}\left [1+\frac{3}{2}\frac{2(\vec{r}-\vec{r'}).\vec{l}}{|\vec{r}-\vec{r'}|^2}+... \right ]\rightarrow 2 \end{align*}

From equation 1 and 2,\begin{align*}E(\vec{r})=\frac{q}{4 \pi \epsilon_0}\left [ (\vec{r}-\vec{r'}-\vec{l}) |\vec{r}-\vec{r'}|^{-3}\left \{ 1+\frac{3(\vec{r}-\vec{r'}).\vec{l}}{|\vec{r}-\vec{r'}|^2}-\frac{(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^3} \right \}\right ] \end{align*}\begin{align*} =\frac{q}{4 \pi \epsilon_0}\left [ \frac{3(\vec{r}-\vec{r'}).\vec{l}(\vec{r}-\vec{r'}-\vec{l})}{|\vec{r}-\vec{r'}|^5}-\frac{\vec{l}}{|\vec{r}-\vec{r'}|^3} \right ]\rightarrow 3\end{align*}

If \(l\rightarrow 0,q\rightarrow \infty \)such that ql remains finite.

\(\therefore \) the dipole moment \(\vec{p}=q\vec{l}, l\rightarrow 0,q\rightarrow \infty\)

Then from 3,\begin{align*}E(\vec{r})=\frac{1}{4 \pi \epsilon_0}\left [ \frac{3(\vec{r}-\vec{r'}).q\vec{l}(\vec{r}-\vec{r'}-\vec{l})}{|\vec{r}-\vec{r'}|^5}-\frac{q\vec{l}}{|\vec{r}-\vec{r'}|^3} \right ] \end{align*}\begin{align*}E(\vec{r})=\frac{1}{4 \pi \epsilon_0}\left [ \frac{3(\vec{r}-\vec{r'}).\vec{p}(\vec{r}-\vec{r'}-\vec{l})}{|\vec{r}-\vec{r'}|^5}-\frac{\vec{p}}{|\vec{r}-\vec{r'}|^3}\right ]\rightarrow 4 \end{align*}

Similarly, the potential due to this dipole may be written as \begin{align*}\phi =V=\frac{1}{4 \pi \epsilon_0}[\frac{1}{|\vec{r}-\vec{r'}-\vec{l}|}-\frac{1}{|\vec{r}-\vec{r'}|}]\rightarrow 5 \end{align*}

By expanding the first term in exactly the same way as was done for equation 2 in above expression and retaining only the linear term in \(\vec{l}\), the equation 5 becomes as \begin{align*}\phi =V=\frac{q}{4 \pi \epsilon_0}\left [ \frac{(\vec{r}-\vec{r'}).\vec{l}}{|\vec{r}-\vec{r'}|^3}\right ] \end{align*}\begin{align*}=\frac{1}{4 \pi \epsilon_0}\left [ \frac{(\vec{r}-\vec{r'}).\vec{p}}{|\vec{r}-\vec{r'}|^3}\right ] \rightarrow 6\end{align*}

From this potential function, the electric field \(\vec{E} \) can be obtained by using relation \begin{align*}E=-grad\phi =-\bigtriangledown \phi \end{align*}

Torque on an Electric Dipole Placed In a Uniform Electric Field

Consider an electric dipole placed in uniform electric field \(\vec{E}\). The dipole consists of charges -q placed at A and +q placed at B. The mid point of AB is O and the length of dipole AB=d. Suppose the axis of the dipole AB makes an angle \(\theta \) with the electric field at a certain instant.

The force on the charge +q is \(\vec{F_1}=q\vec{E}\) and the force on charge -q is\(\vec{F_2}=-q\vec{E}\). Let us calculate the torques (\(\vec{r}\times \vec{F}\)) of these forces about O.

The torque of \(\vec{F_1}\) about O is

\begin{align*}\tau_1=\vec{OB}\times \vec{F_1}=q(\vec{OB}\times \vec{E}) \end{align*}

and the torque of \(\vec{F_2}\) about O is

\begin{align*}\tau_2=\vec{OA}\times \vec{F_2}=-q(\vec{OA}\times \vec{E}) \end{align*}\begin{align*}=q(\vec{AO}\times \vec{E}) \end{align*}

Now the net torque acting on the dipole is \begin{align*}\tau =\tau_1+\tau_2 \end{align*} \begin{align*}= q(\vec{OB}\times \vec{E}) +q(\vec{AO}\times \vec{E})\end{align*}\begin{align*}=q[(\vec{OB}+\vec{AO})\times \vec{E}] =q[\vec{AB}\times \vec{E}]\end{align*}\begin{align*}=q\vec{d}\times \vec{E} \end{align*} \begin{align*}\tau=\vec{p}\times \vec{E} \end{align*}

where \(\vec{p}=q\vec{d} \) is an electric dipole moment.

The direction of torque is perpendicular to the plane containing the dipole axis and the electric field.

\(\therefore \) The magnitude of torque is\begin{align*}\tau =|\vec{\tau}|=PEcos\theta \end{align*}

Potential Energy of a Dipole place in a Uniform Electric Field

We know that an electric dipole placed in a uniform electric field experiences a torque which tends to align it with the field. Here the external field must do certain work to change the orientation of an electtric dipole in external field. This amount if work done is stored as potential energy in the system.

Work done due to the rotation of dipole through an addition angle \(d\theta \) is

\begin{align*}dw=\tau d\theta\end{align*}

Where \(\tau \) is a torque acting on the electric dipole when it is placed in an electric field \(\vec{E}\),\begin{align*}dw=-PEsin\theta d\theta \end{align*}

The work is negative as the rotation \(d\theta \) is opposite to the torque.

The change in electric potential energy of the dipole is therefore \begin{align*}du=-dw=PEsin\theta d\theta \end{align*}

If the dipole is rotating from initial position \(90^{\circ}\) to \(\theta \), the change in P.E. is\begin{align*}U_{\theta}-U_{90^{\circ}}=\int_{90^{\circ}}^{\theta}PEsin\theta d\theta \end{align*}=-PEcos\theta \begin{align*}=-\vec{P}.\vec{E} \end{align*}

If we choose the potential energy of the dipole to be zero when \(\theta =90^{\circ}\) (axis of dipole is perpendicular to field) \(U_{90^{\circ}=0 \) then \begin{align*}U_{\theta}=-\vec{P}.\vec{E} \end{align*}

Potential energy of an Electric Dipole placed in an external field

Let \(E_{ext}(\vec{r})\) be the external electric field, which may be described by the potential function\(V_{ext}(\vec{r})\). If -q and +q charges are placed at \(\vec{r}\) and \(\vec{r}+\vec{l}\) respectively in the external field, then the potential energy U is given by

\begin{align*}U=-q.V_{ext}(\vec{r})+q.V_{ext}(\vec{r}+\vec{l}) \rightarrow 1\end{align*}

From Taylor's series , as l<<<\(\vec{r}\) then \begin{align*}V_{ext}(\vec{r}+\vec{l})=V(\vec{r})+\bigtriangledown V.\vec{l}+... \end{align*}

Hence, the potential energy is \begin{align*}U=q[-V(\vec{r})+V(\vec{r})+\bigtriangledown V.\vec{l}+... ] \end{align*}\begin{align*}=q. [\bigtriangledown V.\vec{l}+... ]\end{align*} \begin{align*}= \bigtriangledown V. q\vec{l}\end{align*}\begin{align*}=\bigtriangledown V.\vec{p}[\because vec{p}=q.vec{l} ] \end{align*}\begin{align*}\therefore U=-\vec{p}.\vec{E}_{ext}(\vec{r}) \end{align*}

References

Adhikari, Pitri Bhakta. A Textbook of Physics Volume-I. Kathmandu: Sukunda Pustak Bhawan, 2015.

Griffiths, D.J., Introduction to Electrodynamics, PHI Learning Private Limited, 2013