Question

An electron on the axis of an electric dipole is$\mathbf{25}\mathbf{\text{nm}}$from the center of the dipole.What is the magnitude of the electrostatic force on the electron if the dipole moment is$\mathbf{3}\mathbf{.6}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{29}}\mathbf{\text{C}}\mathbf{\cdot }\mathbf{\text{m}}$? Assume that$\mathbf{25}\mathbf{\text{nm}}$is much larger than the separation of the charged particles that form the dipole.

Short answer

The magnitude of the electrostatic force on the electron is$6.6\times {10}^{-15}\text{\hspace{0.17em}N}$

Step-by-step solution

The given data

Electron distance from the centre of the dipole,$r=25\text{\hspace{0.17em}nm}$

Dipole moment,$p=3.6\times {10}^{-29}\text{C}\cdot \text{m}$

Electron distance from dipole is larger than the separation of the charged particles.

Understanding the concept of electric field

The magnitude of the net electric field due to a dipole at an axial position for particle distance from dipole being larger than the separation,

$\mathbf{\left|}\overrightarrow{{\mathbf{E}}_{\mathbf{n}\mathbf{e}\mathbf{t}}}\mathbf{\right|}\mathbf{\approx }\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{o}}}\frac{\mathbf{q}\mathbf{d}}{{\mathbf{z}}^{\mathbf{3}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{o}}}\frac{\mathbf{p}}{{\mathbf{z}}^{\mathbf{3}}}$ (i)

where, p = dipole moment

Z = distance of the field point at axial position

Using the concept of the electric field of a dipole, we can get the magnitude of the force using the given data.

Calculation of the magnitude of the electrostatic force

Force on the electron can be given using the given data in equation (i) as follows:

$\begin{array}{c}F=\frac{2kep}{z^3}\\ =\frac{2(8.99\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})(1.60\times {10}^{-19}\text{\hspace{0.17em}C})(3.6\times {10}^{-29}\text{C}\cdot \text{m})}{{(25\times {10}^{-9}\text{\hspace{0.17em}C})}^3}\\ =6.6\times {10}^{-15}\text{\hspace{0.17em}N}\end{array}$

If the dipole is oriented such that is in the +z direction, then $\overrightarrow{F}$ points in the –z direction. Hence, the value of the force is.$6.6\times {10}^{-15}\text{\hspace{0.17em}N}$