Electric Field and Potential Energy of an Electric Dipole

An electric dipole is a pair of charges having equal magnitudes but opposite sign separated at a distance, say $d$. When such a dipole is placed in a uniform electric field, the electric field exerts force on the dipole which then rotates the dipole in clockwise or anticlockwise direction. Here we discuss the electric field and potential energy of an electric dipole.

Electric Field of a Dipole Equidistant from both Charges

Here we determine the electric field of an electric dipole. Consider a positive and a negative charges having equal magnitudes separated at a distance $d$. Such arrangement of charges is called an electric dipole. It's a good idea to start with a coordinate system as shown in Figure 1.

Now we determine the electric field at any point $p$ which is located at the same distance $r$ from both charges. Both charges have the same magnitude so the electric field magnitude at the point $p$ is also the same which is

\[E = k\frac{q}{{{r^2}}}\]

The y-component of electric field due to the electric dipole is a zero vector, that is the y-component of one charge is equal in magnitude and opposite in direction to the y-component of another charge. The y-component of $\vec E_1$ due to positive charge is $E \sin \theta \hat j$ and the y-component of $\vec E_2$ due to negative charge is $-E \sin \theta \hat j$, so they cancel each other. Here the unit vector $\hat j$ is the unit vector along y-axis. Both x-components of electric fields due to the electric dipole lie along the same line (parallel to x-axis) in the same direction and therefore the electric field at the point $p$ is only due to the x-components of electric fields of both charges.

The x-component of electric field due to one charge is $E_x = E \cos \theta$ which is equal in both magnitude and direction to the x-component of electric field of another charge. So the total electric filed at the point $p$ is twice the x-component of electric field due to one charge that is, $E = 2E_x = 2E \cos \theta$. You know the electric field magnitude $E$ from the above equation and therefore, the total electric field is

\[E = k\frac{2q \cos \theta}{r^2} \tag{1} \label{1}\]

In vector form if the unit vector towards x-direction is $\hat i$, the above equation is

\[\vec E = k{\frac{2q \cos \theta}{r^2}} \hat i \tag{2} \label{2}\]

Electric Field of an Electric Dipole on its Axis

Now we find the electric field of an electric dipole at a point on the axis joining the two charges. The Figure 2 shows that the centre of our coordinate system is the centre of the dipole. We are going to find the electric field at the point $p$ shown in Figure 2.

It's quiet simple that you need to add the electric fields due to both charges at the point. Consider that the electric field due to positive charge is $\vec E_1$ and the electric field due to negative charge is $\vec E_2$. In an electric dipole the magnitude of both charges is the same say $q$ and are separated by a distance $d$. Therefore, the distance of positive charge from the point is $y + d/2$ and the distance of negative charge from the point is $y - d/2$. Note that the x-components of electric fields due to both charges is zero. The electric field vectors $\vec E_1$ and $\vec E_2$ are

\[{\vec E_1} = k\frac{q}{{{{\left( {y + \frac{d}{2}} \right)}^2}}} \hat j\]

\[{\vec E_2} = -k\frac{q}{{{{\left( {y - \frac{d}{2}} \right)}^2}}} \hat j\]

The unit vector $\hat j$ gives the direction to the electric field vector which is along y-axis. If $E_{1y}$ is the y-component of $E_1$ and $E_{2y}$ is the y-component of $E_2$, then you know that $E_1 = E_{1y}$ and $E_2 = E_{2y}$ (there is no x-component of electric field at the point $p$). The net electric field which is $\vec E = \vec E_y$ (the subscript y-represents the y-component) at the point $p$ is

\[\begin{align*} \vec E &= k\left[ {\frac{q}{{{{\left( {y + \frac{d}{2}} \right)}^2}}} - \frac{q}{{{{\left( {y - \frac{d}{2}} \right)}^2}}}} \right]\widehat j\\ {\rm{or,}}\quad \vec E &= k\frac{q}{{{y^2}}}\left[ {{{\left( {1 + \frac{d}{{2y}}} \right)}^{ - 2}} - {{\left( {1 - \frac{d}{{2y}}} \right)}^{ - 2}}} \right]\widehat j \end{align*}\]

Note that in an approximation that $y$ is much larger than $d$, the term obviously $\left| \frac{d}{2y} \right| < 1$. Now we use the binomial expansion to solve the terms ${\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}}$ and ${\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}}$. Note that the expression for the binomial expansion of ${(1 + x)^n}$ when $\left| x \right|<1$ is ${{(1+x)}^{n}}=1+nx+n(n-1)\frac{{{x}^{2}}}{2}+...$. So,

\[\begin{align*} {\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}} &= 1 + \frac{d}{y} + \frac{3}{4}\left( {\frac{{{d^2}}}{{{y^2}}}} \right) + ...\\ {\rm{and,}}\quad {\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}} &= 1 - \frac{d}{y} + \frac{3}{4}\left( {\frac{{{d^2}}}{{{y^2}}}} \right) + ... \end{align*}\]

Now keeping only the first two terms neglecting the smaller terms we have ${\left( {1 - \frac{d}{{2y}}} \right)^{ - 2}} \cong 1 + \frac{d}{y}$ and ${\left( {1 + \frac{d}{{2y}}} \right)^{ - 2}} \cong 1 - \frac{d}{y}$. So the net electric field is,

\[\begin{align*} \vec E &= k\frac{q}{{{y^2}}}\left[ {\left( {1 - \frac{d}{y}} \right) - \left( {1 + \frac{d}{y}} \right)} \right]\hat j\\ &= - k\frac{{2qd}}{{{y^3}}}\hat j = - k\frac{{2p}}{{{y^3}}}\hat j = k\frac{{2\vec p}}{{{y^3}}} \tag{3} \label{3} \end{align*}\]

As you can see from the above expression of the net electric field that the electric field is proportional to $\frac{1}{{{y^3}}}$ instead of $\frac{1}{{{y}^{2}}}$. The above expression of net electric field tells us that the net electric field is along negative y-direction in our case shown in Figure 2. If you consider only the magnitude of the net electric field, it is

\[E = k\frac{2p}{y^3} \tag{4} \label{4}\]

Potential Energy of an Electric Dipole

Here we find the potential energy of an electric dipole in a uniform electric field. Consider that the dipole is inside a uniform electric field as shown in Figure 3. The two charges of the dipole are separated at a distance $d$. The dipole makes an angle $\theta $ with the direction of electric field. The electric field exerts force on each charge of the dipole. So the torque produced tends to rotate the dipole in anticlockwise direction.

The force on negative charge is $F_1$ and on positive charge is $F_2$. The magnitude of force on each charge is the same. Let the magnitude of one charge is $q$ and therefore the magnitude of force on each charge is $F = qE$ where $E$ is the electric field magnitude. The perpendicular distance between the line of action of forces (shown in dotted line in Figure 3) is $d\sin \theta $ so the lever arm for each force is the same which is $\frac{d}{2}\sin \theta $. The magnitude of torque $\tau $ for each charge is also the same which is $(qE)\left( \frac{d}{2}\sin \theta \right)$. Since both torques tend to rotate the dipole in anticlockwise direction, the net torque magnitude on the dipole is twice the torque magnitude on one of the charges which is:

\[\tau = qdE\sin \theta {\rm{ }} \tag{5} \label{5}\]

The product $qd$ is another physical quantity called electric dipole moment. The electric dipole moment $\vec{p}$ has a direction from negative charge to positive charge in an electric dipole. Now in terms of the electric dipole moment, the above expression can be written as

\[\tau = pE\sin \theta \tag{6} \label{6}\]

This is the expression for the cross product of vectors, so in vector form it is $\vec{\tau }=\vec p \times \vec E$. You need to know the right hand thumb rule of vector product to know the direction of $\vec \tau$; the curved fingers give the direction of rotation and the thumb gives the direction of $\vec \tau$ which in this case is perpendicularly towards you. As you can see in Figure 3 and from above equation the torque is zero when $\theta $ is zero or $\pi $. When $\theta =0$, $\vec p$ and $\vec E$ are antiparallel which is the position of unstable equilibrium. When $\theta =\pi $, $\vec p$ and $\vec E$ are parallel which is the position of stable equilibrium. And the torque always tends to rotate the dipole in stable equilibrium position.

Now let the torque rotates the dipole through a small angle $d\theta $ , so the small work done by the torque is $dW=\tau d\theta $. Since the torque rotates the dipole in anticlockwise direction, that is in the direction of increasing $\theta $ the work done is positive. The total work done by the torque is obtained by integrating $dW$ between limits $\theta_1$ and $\theta_2$:

\[W = \int\limits_{{\theta _1}}^{{\theta _2}} {\tau d\theta } = pE\int\limits_{{\theta _1}}^{{\theta _2}} {\sin \theta {\mkern 1mu} d\theta } = pE( - cos{\theta _2} + cos{\theta _1})\]

\[{\rm{or,}}\quad W = pE\cos {\theta _1} - pE\cos {\theta _2} \tag{7} \label{7}\]

In the above equation Eq. \eqref{7}, the quantity $pE \cos \theta$ is the potential energy of the electric dipole. It is denoted by $U$ and therefore, $U_1 = pE \cos \theta_1$ and $U_2 = pE \cos \theta_2$. So, $W=U_1 - U_2 = -(U_2 - U_1) = -\Delta U$. Therefore work done is the negative of change in potential energy. You know from the conservation of mechanical energy that the work done by gravitational force is also the negative of change in gravitational potential energy. The potential energy of the electric dipole is

\[U = pE \cos \theta \label{8} \tag{8}\]

In anticlockwise direction $\theta $ increases and the potential energy goes on decreasing until becomes minimum in stable equilibrium position at $\theta = \pi$. At $\theta = \pi$, the potential energy is $U = -pE$ which is the most negative value. In anticlockwise direction the work done is positive; final potential energy is smaller than initial potential energy ($U_2 < U_1$) and the negative of change in potential energy is positive. Similarly in clockwise direction that is in the direction of decreasing $\theta $ the work done is negative; final potential energy is greater than initial ($U_2 > U_1$) and the negative of change in potential energy is negative. Note that the torque tends to minimize the potential energy of the dipole towards stable equilibrium position. At $\theta = 0$ the potential energy is maximum which is $U = pE$ and zero at $\theta = \pi /2$. Note that zero potential energy does not mean that the the dipole does not have potential energy but you know that zero is greater than negative values.

If the torque rotates the dipole in clockwise direction (the electric field direction should be exactly opposite to the direction shown in Figure 3) which is in the direction of decreasing $\theta $, the work done should be positive (the torque is in the same direction of rotation). In this case the final potential energy is greater than initial and therefore the potential energy of the dipole is $U=-pE\cos \theta $.