Question

An electron is released from rest on the axis of an electric dipole that has charge eand charge separation $\mathbf{d}\mathbf{=}\mathbf{20}\mathbf{\text{\hspace{0.17em}pm}}$pm and that is fixed in place. The release point is on the positive side of the dipole, at distance$\mathbf{\text{7.0d}}$from the dipole center. What is the electron’s speed when it reaches a point$\mathbf{\text{5.0d}}$from the dipole center?

Short answer

$\mathrm{v}=7.0\times {10}^5\text{\hspace{0.17em}m/s}$

Step-by-step solution

Given

$\mathrm{d}=20\text{\hspace{0.17em}pm}=20\times {10}^{-12}\text{\hspace{0.17em}m}$

Understanding the concept

After reading the question, The electric potential energy in the presence of a dipole is$\mathrm{U}={\mathrm{qV}}_{\mathrm{dipole}}$.

$\begin{array}{c}\mathrm{U}=\frac{\mathrm{qp}\text{\hspace{0.17em}cos\hspace{0.17em}}\mathrm{\theta }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\\ \mathrm{U}=\frac{(-\mathrm{e})\left(\mathrm{ed}\right)\text{cos\hspace{0.17em}}\mathrm{\theta }}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{r}}^2}\\ {\mathrm{\theta }}_{\mathrm{i}}={\mathrm{\theta }}_{\mathrm{f}}=0^{\circ }.\end{array}$

Apply the Law of Conservation of Energy: ${\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{i}}$.

Calculate the electron’s speed when it reaches a point  5.0d from the dipole center

${\mathrm{K}}_{\mathrm{f}}+{\mathrm{U}}_{\mathrm{f}}={\mathrm{K}}_{\mathrm{i}}+{\mathrm{U}}_{\mathrm{i}}$

$\begin{array}{c}\frac{1}{2}{\mathrm{mv}}^2=\frac{{\mathrm{e}}^2\mathrm{d}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{d}}^2}\left(\frac{1}{5^2}-\frac{1}{7^2}\right)\\ =\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{d}}\left(\frac{1}{25}-\frac{1}{49}\right)\\ {\mathrm{v}}^2=\frac{2{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{dm}}\left(\frac{1}{25}-\frac{1}{49}\right)\mathrm{d}=20\times {10}^{-12}\text{\hspace{0.17em}m},\mathrm{m}=9.11\times {10}^{-31}\text{\hspace{0.17em}kg}\\ \mathrm{v}=\sqrt{\frac{2{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{dm}}\left(\frac{1}{25}-\frac{1}{49}\right)}\\ \mathrm{v}=7.0\times {10}^5\text{\hspace{0.17em}m/s}\end{array}$

Thereforethe electron’s speed is$\mathrm{v}=7.0\times {10}^5\text{\hspace{0.17em}m/s}$