Q45P - Fundamentals Of Physics

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

 

Step 3: 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.$6.6\times {10}^{-15}\text{\hspace{0.17em}N}$