91Ó°ÊÓ

Electron distance from the centre of the dipole,$r=25\text{ n³¾}$

Dipole moment,$p=3.6×{10}^{−29}\text{C}â‹\dots \text{m}$

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

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,

$\mathbf{\left|}\stackrel{\mathbf{→}}{{\mathbf{E}}_{\mathbf{n}\mathbf{e}\mathbf{t}}}\mathbf{\right|}\mathbf{≈}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{o}}}\frac{\mathbf{q}\mathbf{d}}{{\mathbf{z}}^{\mathbf{3}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{o}}}\frac{\mathbf{p}}{{\mathbf{z}}^{\mathbf{3}}}$ (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.

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

$\begin{array}{c}F=\frac{2kep}{z^3}\\ =\frac{2(8.99×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})(1.60×{10}^{−19}\text{ C})(3.6×{10}^{−29}\text{C}â‹\dots \text{m})}{{(25×{10}^{−9}\text{ C})}^3}\\ =6.6×{10}^{−15}\text{ N}\end{array}$

If the dipole is oriented such that is in the +z direction, then $\stackrel{→}{F}$ points in the –z direction. Hence, the value of the force is.$6.6×{10}^{−15}\text{ N}$