An electron (mass = $\mathbf{9}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{kg}$) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm away. It reaches the grid with a speed of . If the accelerating force is constant, compute

(a) the acceleration;

(b) the time to reach the grid; and

(c) the net force, in newtons. Ignore the gravitational force on the electron.

The mass of the electron is

$m=9.11\times {10}^{-31}$

The initial velocity of the electron is

u = 0 m/s

The final velocity of the electron is

$v=3\times {10}^6\mathrm{m}/\mathrm{s}$

The distance traveled by the electron is

$$\begin{gathered} S=1.8\mathrm{cm} \\ =1.8.\left(1\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}\right) \\ =0.018\mathrm{m} \end{gathered}$$

The initial velocity u , final velocity v , acceleration a and the distance traveled S are related as

${\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{aS}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

The initial velocity u, final velocity v , acceleration a and the time of travel t are related as

$\mathbf{v}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{at}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

According to the second law of motion, the force on an object of mass m and acceleration a is

F = ma ........(3)

Let the acceleration of the electron be a. From equation (1),

$$\begin{gathered} \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{S}} \\ =\frac{{\left(3\times {10}^6\mathrm{m}/\mathrm{s}\right)}^2-0^2}{2\times 0.018\mathrm{m}} \\ =2.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the accelaration is$2.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2$

Let the time of motion of the electron be t . From equation (2),

$$\begin{gathered} \mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}} \\ =\frac{3\times {10}^6\mathrm{m}/\mathrm{s}-0}{2.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2} \\ =1.2\times {10}^{-8}\mathrm{s} \end{gathered}$$

Thus, the time of motion of the electron is $1.2\times {10}^{-8}\mathrm{s}$.

From equation (3), the force on the electron is

$$\begin{gathered} F=ma \\ =9.11\times {10}^{-31}\mathrm{kg}\times 2.5\times {10}^{14}\mathrm{m}/{\mathrm{s}}^2 \\ =22.78\times {10}^{-17}\mathrm{N} \end{gathered}$$

Thus, the force on the electron is $22.78\times {10}^{-17}\mathrm{N}$.