Consider a dipole with charges $q_1 = +q$ and $q_2= –q$ placed in a uniform electric field E, as shown in Fig. 2.16.
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| Fig.1: Potential energy of a dipole in a uniform external field. |
As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque τ given by
τ $\mathbf{=p \times E}$ (1)
which will tend to rotate it (unless $\mathbf{p}$ is parallel or antiparallel to $\mathbf{E}$). Suppose an external torque $\mathbf{\tau_{ext}}$ is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle $\theta_0$ to angle $\theta_1$ at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by
$W=\int_{\theta_0}^{\theta_1}t_{ext}d\theta=\int_{\theta_0}^{\theta_1}pE \ sin \ \theta d\theta$
$W=pE(cos \ \theta_0-cos \ \theta_1)$ (2)
This work is stored as the potential energy of the system. We can then associate potential energy $U(\theta)$ with an inclination $\theta$ of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy $U$ is taken to be zero. A natural choice is to take $\theta_0 = \pi/2$. (Αn explanation for it is provided towards the end of discussion.) We can then write,
$U(\theta)=pE\left(cos \ \frac{\pi}{2}-cos \ \theta\right)=pE \ cos \ \theta=-\mathbf{p.E}$ (3)
This expression can alternately be understood also from Eq. (2.29). We apply Eq. (2.29) to the present system of two charges $+q$ and $–q$. The potential energy expression then reads
$U'{\theta}=q[V(\mathbf{r_1})-V(\mathbf{r_2})]-\frac{q^2}{4\pi \epsilon_0 \times 2a}$ (4)
Here, $r_1$ and $r_2$ denote the position vectors of $+q$ and $–q$. Now, the potential difference between positions $r_1$ and $r_2$ equals the work done in bringing a unit positive charge against field from $r_2$ and $r_1$. The displacement parallel to the force is $2a \ cos \ \theta$. Thus, $[V(\mathbf{r_1})-V(\mathbf{r_2})] = - E \times 2a \ cos \ \theta$ . We thus obtain,
$U'{\theta}=-pE \ cos \ \theta- \frac{q^2}{4\pi \epsilon_0 \times 2a}=-\mathbf{p.E}-\frac{q^2}{4\pi \epsilon_0 \times 2a}$
We note that $U′ (\theta)$ differs from $U(\theta)$ by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Eq. (2.34) and it then reduces to Eq. (2.32).
We can now understand why we took $\theta_0 =\pi/2$. In this case, the work done against the external field $\mathbf{E}$ in bringing $+q$ and $–q$ are equal and opposite and cancel out, i.e., $q[V(\mathbf{r_1})-V(\mathbf{r_2})]=0$.
Significance of Dipole in External Field
The significance of a dipole in an external electric field lies in its ability to interact with the field and experience a torque and potential energy that depend on its orientation relative to the field.
- When placed in an external electric field, a dipole tends to align itself with the field. If the dipole moment vector is parallel to the electric field, the system is at a stable equilibrium, with minimum potential energy.
- The dipole experiences a torque in the presence of an external electric field, which tends to rotate the dipole to align it with the field.
- The dipole also possesses potential energy when placed in an external electric field. This potential energy depends on the orientation of the dipole relative to the field and is minimized when the dipole aligns with the field.
- The behavior of dipoles in external fields is used in in determining molecular structure, reactivity, and solvation.
Applications:
- The interaction of dipoles with external electric fields is significant in:
- Molecular Physics: Understanding molecular polarity, structure, and interactions.
- Reactivity and Solvation: Explaining how molecules interact with solvents and other reactive species.
- Dielectrics: The behavior of insulating materials in electric fields.
- Technology: Applications in sensors, actuators, and other devices relying on dipole-field interactions.
Example 1
A molecule of a substance has a permanent electric dipole moment of magnitude 10–29 C m. A mole of this substance is polarised (at low temperature) by applying a strong electrostatic field of magnitude 106 V/m. The direction of the field is suddenly changed by an angle of $60^0$. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarisation of the sample.
Solution:
Here, dipole moment of each molecules $= 10^{–29} C.m$. As 1 mole of the substance contains $6 \times 10^{23}$ molecules, total dipole moment of all the molecules,
$p=6 \times 10^{23} \times 10^{-29} \ C.m$
$p=6 \times 10^{-6} \ C.m$
Initial potential energy,
$U_i = –pE cos \ \theta = –6 \times 10^{–6}\times 10^6 \ cos \ 0° = –6 \ J$
Final potential energy (when $\theta = 60^0$),
$U_f = –pE cos \ \theta = –6 \times 10^{–6}\times 10^6 \ cos \ 60° = –3 \ J$
Change in potential energy $= –3 \ J – (–6 \ J) = 3 \ J$
So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.
Example 2
A water molecule has an electric dipole moment of $6.3 \times 10^{-30} \ C.m$. A sample contains $10^{22}$ water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude $3 \times 10^5 N.C^{-1}$. How much work is required to rotate all the water molecules from $\theta = 0º$ to $90º$?
Solution
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from $θ = 0º$ to $90º$ is equal to the potential energy difference between these two configurations.
$W= \Delta U =U (90º)−U (0º)$
From the equation (1.51), we write $U = − pE cos \ \theta$, Next we calculate the work done to rotate one water molecule from $\theta = 0º$ to $90º$.
For one water molecule
$W = −pE cos 90º + pE cos0º = pE$
$W = 6.3 \times 10^{−30} \times 3 \times 10^5 = 18 .9 \times 10^{−25} \ J$
For $10^{22}$ water molecules, the total work done is
$W_{tot} = 18.9 \times 10^{−25} \times 10^{22} =18.9 \times 10^{−3} \ J$
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