Question

Electron in a well. Figure 24- 65 shows electric potential $\mathbf{V}$ along an x-axis. The scale of the vertical axis is set by${\mathbf{V}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{8.0}\mathbf{\text{V}}$ . An electron is to be released at $\mathbf{x}\mathbf{}\mathbf{=}\mathbf{4}\mathbf{.5}\mathbf{\text{cm}}$with initial kinetic energy $\mathbf{\text{3.00eV}}$. (a) If it is initially moving in the negative direction of the axis, does it reach a turning point (if so, what is the x coordinate of that point) or does it escape from the plotted region (if so, what is its speed atrole="math" localid="1662990552191" $\mathbf{\text{x=0}}$ )? (b) If it is initially moving in the positive direction of the axis, does it reach a turning point (if so, what is the x coordinate of that point) or does it escape from the plotted region (if so, what is its speed at $\mathbf{\text{x=7.0cm}}$)? What are the (c) magnitude $\mathbf{\text{F}}$ and (d) direction (positive or negative direction of the x axis) of the electric force on the electron if the electron moves just to the left of$\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}\mathbf{\text{cm}}$ ? What are (e) $\mathbf{\text{F}}$ and (f) the direction if it moves just to the right of$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5.0}\mathbf{\text{cm}}$ ?

Short answer

1. The$\mathrm{x}$-coordinate of that point is,$\mathrm{x}=1.8\text{cm}$.
2. The speed of the electron is,$8.4\times {10}^5\text{m}/\text{s}$.
3. The force on the electron is,$\mathrm{F}=2.13\times {10}^{-17}\mathrm{N}$.
4. The force on the electron is along the positive-direction.
5. Force on the electron is,$\mathrm{F}=-1.6\times {10}^{-17}\text{N}$.
6. The direction of the force on the electron is along the negative $\mathrm{x}$-direction.

Step-by-step solution

Step 1: Given data:

The charge on the electron,${\mathrm{q}}_{\mathrm{e}}=-1.6\times {10}^{-19}\text{C}$

The mass of the electron,${\mathrm{m}}_{\mathrm{e}}=9.11\times {10}^{-31}\text{kg}$

The potential at$\mathrm{x}=\text{4.5cm}$, ${\mathrm{V}}_1=\text{6.00V}$

Therefore, the potential energy of the electron at$\mathrm{x}=4.5\text{cm}$, ${\mathrm{U}}_1=-6.00\text{eV}$

The kinetic energy of the electron at$\mathrm{x}=4.5\text{cm}$, ${\mathrm{K}}_1=3.00\text{eV}$

Hence, the total energy of the electron is,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{\tau }}=-6.00\text{eV}+3.00\text{eV}\\ =-3.00\text{eV}\end{array}$

Determining the concept:

Use the formula for force acting on an electron and the kinetic energy. Observe the graph to conclude the direction of force acting on electron.

Formulae are as follow:

The kinetic energy is,

${\mathrm{K}}_2={\mathrm{E}}_{\mathrm{T}}-{\mathrm{U}}_2$

Here, ${\mathrm{E}}_{\mathrm{T}}$ is the total energy and ${\mathrm{U}}_2$ is the potential energy.

The velocity is,

$\mathrm{v}=\sqrt{\frac{2{\mathrm{K}}_2}{{\mathrm{m}}_{\mathrm{e}}}}$

Here, ${\mathrm{m}}_{\mathrm{e}}$is the mass of electron.

The force is,

$\mathrm{F}={\mathrm{q}}_{\mathrm{e}}\mathrm{E}$

Where, $\mathrm{K}$ is kinetic energy, $\mathrm{E}$ is total energy, $\mathrm{U}$ is potential energy, $\mathrm{v}$ is velocity, $\mathrm{F}$ is force , and $\mathrm{q}$ is charge on electron.

(a) Determining if the electron reaches a turning point:

If the electron reaches a turning point where it's kinetic energy ${\mathrm{K}}_2=0$ towards the left, so its potential energy should be $-3.00\text{eV}$.

Therefore, the potential is $\text{3.00V}$.

From the graph, the potential is$\text{3.00V}$when it reaches$\mathrm{x}=1.8\text{cm}$.

So, the electron reaches the turning point.

Therefore, the $\mathrm{x}$-coordinate of that point is $\mathrm{x}=1.8\text{cm}$.

(b) Determining the speed of the electron:

If the electron is initially moving in the positive direction of the axis, then it does not reach the turning point.

So, the electron will escape from the plotted region.

From above graph, the potential at $\mathrm{x}=7.0\text{cm}$is $5.00\text{V}$.

So, at this point the potential energy of the electron ${\mathrm{U}}_2=-5.00\text{eV}$.

Therefore, the kinetic energy of the electron is,

$\begin{array}{c}{\mathrm{K}}_2={\mathrm{E}}_{\mathrm{T}}-{\mathrm{U}}_2\\ =-3.00\text{eV}-(-5.00\text{eV})\\ =\left(\text{2.00eV}\right)\left(\frac{\text{1}.6\times {10}^{-19}\text{J}}{\text{1eV}}\right)\\ =3.2\times {10}^{-19}\text{J}\end{array}$

The kinetic energy of the electron is given by ${\mathrm{K}}_2=\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2$ .

Therefore, the speed of the electron is,

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{2{\mathrm{K}}_2}{{\mathrm{m}}_{\mathrm{e}}}}\\ =\sqrt{\frac{\text{2}(3.2\times {10}^{\text{-19}}\text{J})}{(\text{9.11}\times {10}^{\text{-31}}\text{kg})}}\\ =8.4\times {10}^5\text{m}/\text{s}\end{array}$

Hence, the speed of the electron is, $8.4\times {10}^5\text{m}/\text{s}$

(c) Determining the force on the electron:

The slope of the line at the left of $\mathrm{x}=4.0\text{cm}$is,

$\begin{array}{c}\frac{\mathrm{dV}}{\mathrm{dx}}=\frac{(\text{6.00V}-\text{2.00V})}{(\text{4.0cm}-\text{1.0cm})}\\ =\frac{\text{4.00V}}{\left(\text{3.0cm}\right)\left(\frac{{\text{10}}^{\text{-2}}\text{m}}{\text{1cm}}\right)}\\ =\text{133.33V}/\text{m}\end{array}$

Therefore, the electric field is given by,

$\begin{array}{c}\mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dx}}\\ =-133.33\text{V}/\text{m}\end{array}$

Hence, the force on the electron is given by,

$\begin{array}{c}\mathrm{F}={\mathrm{q}}_{\mathrm{e}}\mathrm{E}\\ =(-1.6\times {10}^{-19}\text{C})(-133.33\text{V}/\text{m})\\ =2.13\times {10}^{-17}\text{N}\end{array}$

Hence, the force on the electron is $2.13\times {10}^{-17}\text{N}$.

(d) Determining the direction of the force on the electron:

The value of the force on the electron is greater than zero.

Therefore, the direction of the force on the electron is along the positive$\mathrm{x}$-direction.

(e) Determining the force on the electron:

The slope of the line at the right of $\mathrm{x}=5.0\text{cm}$ is,

$\begin{array}{c}\frac{\mathrm{dV}}{\mathrm{dx}}=\frac{(6.00\text{V}-5.00\text{V})}{(5.0\text{cm}-6.0\text{cm})}\\ =\frac{1.00\text{V}}{(-1.0\text{cm})\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)}\\ =-100\text{V}/\text{m}\end{array}$

Therefore, the electric field is given by,

$\begin{array}{c}\mathrm{E}=-\frac{\mathrm{dV}}{\mathrm{dx}}\\ =-100\text{V}/\text{m}\end{array}$

Hence, the force on the electron is given by,

$\begin{array}{c}\mathrm{F}={\mathrm{q}}_{\mathrm{e}}\mathrm{E}\\ =(-\text{1}.6\times 1{\text{0}}^{\text{-19}}\text{C})(\text{100V}/\text{m})\\ =-1.\text{6}\times 1{\text{0}}^{\text{-17}}\text{N}\end{array}$

(f) Determining the direction of the force on the electron:

The value of the force on the electron is less than zero.

Therefore, the direction of the force on the electron is along the negative$\mathrm{x}$-direction.

Hence, by using the concept of electric force acting on electron, calculation can be done easily.