91影视

• Initial speed of the electron,${\mathrm{v}}_{\mathrm{i}}=5脳{10}^8\text{鈥塩尘/蝉}$
• Final speed of the electron,${\mathrm{v}}_{\mathrm{f}}=0\text{鈥尘/s}$
• Magnitude of electric field,$\left|\stackrel{鈫�}{\mathrm{E}}\right|=1脳{10}^3\text{鈥塏/颁}$
• Length of the region, $\mathrm{L}=8\text{鈥尘尘}$

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,${{\mathrm{v}}_{\mathrm{f}}}^2鈭�{{\mathrm{v}}_{\mathrm{i}}}^2=2\mathrm{补螖虫}$ (ii)

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

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

The force on a body due to Newton鈥檚 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:

$\begin{array}{c}\stackrel{鈫�}{\mathrm{a}}=鈭�\mathrm{e}\stackrel{鈫�}{\mathrm{E}}/{\mathrm{m}}_{\mathrm{e}}\\ =\frac{鈭�1.6脳{10}^{鈭�19}\text{鈥嬧赌塁}脳1脳{10}^3\text{鈥塏/颁}}{9.11脳{10}^{鈭�31}\text{鈥塳驳}}\\ =鈭�1.7582脳{10}^{14}{\text{鈥尘/s}}^{\text{2}}\end{array}$ to solve for distance 鈭唜:

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

$\begin{array}{c}\mathrm{螖虫}=\frac{鈭�{\mathrm{v}}_{\mathrm{i}}^2}{2\mathrm{a}}\\ =\frac{鈭�{(5.00脳{10}^6\text{m/s})}^2}{鈭�2脳1.7582脳{10}^{14}{\text{鈥尘/s}}^{\text{2}}}\\ =7.12脳{10}^{鈭�2}\text{鈥尘}\end{array}$

Hence, the value of the distance is$7.12脳{10}^{鈭�2}\text{鈥尘}$

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

$\begin{array}{c}t=\frac{2螖x}{v_i}\text{(}鈭�v_i=2v_{avg}\text{,蹿辞谤鈥�}{v}_f=0\text{鈥尘/s)}\\ =\frac{2脳7.12脳{10}^{鈭�2}\text{m}}{5.00脳{10}^6{\text{m/s}}^2}\\ =2.84脳{10}^{鈭�8}\text{鈥塻}\end{array}$

Hence, the value of the elapsed time is$2.84脳{10}^{鈭�8}\text{鈥塻}$

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

$\begin{array}{c}\frac{\mathrm{螖碍}}{{\mathrm{K}}_{\mathrm{i}}}=\frac{\mathrm{螖}\left(\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}^2\right)}{\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{\mathrm{螖}\left({\mathrm{v}}^2\right)}{{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{2\mathrm{补螖虫}}{{\mathrm{v}}_{\mathrm{i}}^2}\\ =\frac{2脳1.7582脳{10}^{14}脳7.12脳{10}^{鈭�3}\mathrm{m}}{{(5脳{10}^6\mathrm{m}/\mathrm{s})}^2}\\ =0.10015\\ 鈮�10\mathrm{\%}\end{array}$

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