Figure 24-56ashows an electron moving along an electric dipole axis toward the negative side of the dipole. The dipole is fixed in place. The electron was initially very far from the dipole, with kinetic energy$\mathbf{100}\mathbf{}\mathbf{\text{eV}}$. Figure 24-56bgives the kinetic energy Kof the electron versus its distance rfrom the dipole center. The scale of the horizontal axis is set by${\mathbf{r}}_{\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.10}\mathbf{}\mathbf{\text{m}}$.What is the magnitudeof the dipole moment?

1. Kinetic energy of the dipole,$\mathrm{KE}=100\text{eV}$
2. The scale of the horizontal axis,${\mathrm{r}}_{\mathrm{s}}=0.1\text{m}$

Using the basic concept of potential energy and dipole, we get the equation of the change in the potential energy of a system. Again, using the work-energy concept, we can say that the change in kinetic energy is equal to that of the change in the potential energy in this situation. Hence, using this concept and the given graph data, we can calculate the required value of the dipole moment.

Formulae:

The electric potential due to an electric dipole is,$\mathrm{V}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{pcos\theta }}{{\mathrm{r}}^2}\right)$ (i)

Here,$\frac{1}{4{\mathrm{\pi \epsilon }}_0}$is the coulomb’s constant,$p$ is the dipole momentum of the electric dipole,$\mathrm{r}$ is the distance from the center of the dipole, and$\mathrm{\theta }$ is the angle.

Using the work-energy theorem and the change in kinetic energy is equal to the change in the potential energy. $\mathrm{\Delta K}=\mathrm{\Delta U}$ (ii)

The change in the potential energy for the electric dipole, $\mathrm{\Delta U}=-\mathrm{eV}$ (iii)

Using equation (i) in equation (iii), we get the change in potential energy as:

$\mathrm{\Delta U}=-\frac{\mathrm{e}}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{pcos\theta }}{{\mathrm{r}}^2}\right)$

Again, using equation (ii), we can get the change in kinetic energy of the system as follows:

$\mathrm{\Delta K}=-\mathrm{e}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{pcos\theta }}{{\mathrm{r}}^2}\right)\right)$

The horizontal length given in the figure (b) is${\mathrm{r}}_{\mathrm{s}}=5\mathrm{r}$

Here, $\mathrm{r}$is the one unit in the horizontal axis or distance from the center of the dipole.

Substitute $0.10\mathrm{m}$for ${\mathrm{r}}_{\mathrm{s}}$in the above equation and we get the value of$\mathrm{r}$as follows:

$\begin{array}{c}0.10\text{m}=5\mathrm{r}\\ \mathrm{r}=\frac{0.10\text{m}}{5}\\ =0.020\text{m}\end{array}$

Since the electron is moving along the electric dipole axis toward the negative side of the dipole, the angle$\theta$must be equal torole="math" localid="1661921439037" $108°$

Rearranging the above equation $\mathrm{\Delta K}=-\mathrm{e}\left(\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{\mathrm{pcos\theta }}{{\mathrm{r}}^2}\right)\right)$ for dipole moment, p, and substituting the given data, we get that ( $108°$for $\theta$, $8.854\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2$for ${\mathrm{\epsilon }}_0$, 100eV for $∆k$, and $0.020\mathrm{m}$for $\mathrm{r}$)

$\begin{array}{c}\mathrm{p}=-\frac{{\mathrm{\Delta kr}}^2\left(4{\mathrm{\pi \epsilon }}_0\right)}{\mathrm{ecos\theta }}\\ =\frac{(100\text{eV})\left(\frac{1.6\times {10}^{-19}\text{J}}{1\text{eV}}\right){(0.020\text{m})}^2\left(4\mathrm{\pi }\right)(8.854\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/N.m}}^{\text{2}})}{(1.6\times {10}^{-16}\text{C})\cos {180}^0}\\ =4.5\times {10}^{-12}\text{C.m}\end{array}$

Therefore, the magnitude of the dipole moment is $4.5\times {10}^{-12}\text{C.m}$