An electron with a speed of $\mathbf{5}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{8}}\mathbf{\text{cm/s}}$ enters an electric field of magnitude$\mathbf{1}\mathbf{.00}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{\text{N/C}}$ , traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is$\mathbf{8}\mathbf{.00}\mathbf{\text{\hspace{0.17em}mm}}$ long (too short for the electron to stop within it), what fraction of the electron’s initial kinetic energy will be lost in that region?

• Initial speed of the electron,${\mathrm{v}}_{\mathrm{i}}=5\times {10}^8\text{\hspace{0.17em}cm/s}$
• Final speed of the electron,${\mathrm{v}}_{\mathrm{f}}=0\text{\hspace{0.17em}m/s}$
• Magnitude of electric field,$\left|\overrightarrow{\mathrm{E}}\right|=1\times {10}^3\text{\hspace{0.17em}N/C}$
• Length of the region, $\mathrm{L}=8\text{\hspace{0.17em}mm}$

Using the basic equations of kinematics and Newton's law of motion, the required quantities can be found. Again, this Newtonian force can be related to the electric field using charge.

Formulae:

The force relation to the electric field of a body,$\mathrm{F}=\mathrm{qE}$ (i)

The third law of kinematic motion,${{\mathrm{v}}_{\mathrm{f}}}^2-{{\mathrm{v}}_{\mathrm{i}}}^2=2\mathrm{a\Delta x}$ (ii)

The elapsed time of a motion,$\mathrm{t}=\frac{\mathrm{\Delta x}}{{\mathrm{v}}_{\mathrm{avg}}}$ (iii)

The kinetic energy of a motion,$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$ (iv)

The force on a body due to Newton’s second law of motion, $\mathrm{F}=\mathrm{ma}$ (v)

The initial direction of motion is taken to be the +x direction (this is also the direction of E. again, from the given data, we know that${\mathrm{v}}_{\mathrm{f}}=0\mathrm{m}/\mathrm{s}$

Thus, using equation (i) in equation (v), we can get the acceleration of the electron particle as given:

$\begin{array}{c}\overrightarrow{\mathrm{a}}=-\mathrm{e}\overrightarrow{\mathrm{E}}/{\mathrm{m}}_{\mathrm{e}}\\ =\frac{-1.6\times {10}^{-19}\text{\hspace{0.17em}C}\times 1\times {10}^3\text{\hspace{0.17em}N/C}}{9.11\times {10}^{-31}\text{\hspace{0.17em}kg}}\\ =-1.7582\times {10}^{14}{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$ to solve for distance ∆x:

Now, the distance travelled by the electron after retardation is given using equation (ii) as:

$\begin{array}{c}\mathrm{\Delta x}=\frac{-{\mathrm{v}}_{\mathrm{i}}^2}{2\mathrm{a}}\\ =\frac{-{(5.00\times {10}^6\text{m/s})}^2}{-2\times 1.7582\times {10}^{14}{\text{\hspace{0.17em}m/s}}^{\text{2}}}\\ =7.12\times {10}^{-2}\text{\hspace{0.17em}m}\end{array}$

Hence, the value of the distance is$7.12\times {10}^{-2}\text{\hspace{0.17em}m}$

Using the above value of distance in equation (iii), we can get the elapsed time of the electron’s motion as follows:

$\begin{array}{c}t=\frac{2\Delta x}{v_i}\text{(}\because v_i=2v_{avg}\text{,for\hspace{0.17em}}{v}_f=0\text{\hspace{0.17em}m/s)}\\ =\frac{2\times 7.12\times {10}^{-2}\text{m}}{5.00\times {10}^6{\text{m/s}}^2}\\ =2.84\times {10}^{-8}\text{\hspace{0.17em}s}\end{array}$

Hence, the value of the elapsed time is$2.84\times {10}^{-8}\text{\hspace{0.17em}s}$

Using equation (iv), the fractional value lost in the initial kinetic energy can be given as:

$\begin{array}{c}\frac{\mathrm{\Delta K}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{\mathrm{\Delta }\left(\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right)}{\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{\mathrm{\Delta }\left({\mathrm{v}}^2\right)}{{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{2\mathrm{a\Delta x}}{{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{2\times 1.7582\times {10}^{14}\times 7.12\times {10}^{-3}\mathrm{m}}{{(5\times {10}^6\mathrm{m}/\mathrm{s})}^2}\\ =0.10015\\ \approx 10\mathrm{\%}\end{array}$

Thus, the fraction of the initial kinetic energy lost in the region is $0.1\text{or10\%}$