91Ó°ÊÓ

The mass of the electron is

$m=9.11×{10}^{-31}kg$

The initial velocity of the electron is

u = 0 m/s

The final velocity of the electron is

$v=3×{10}^{6}m/s$

The distance traveled by the electron is

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

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

${\mathit{V}}^{\mathbf{2}}\mathbf{=}{\mathit{u}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{s}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

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

$\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$

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

$\mathit{F}\mathbf{=}\mathit{m}\mathit{a}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right)$

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

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

Thus, the acceleration is $2.5×{10}^{14}m/s^2$.

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

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

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

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

$$\begin{gathered} F=ma \\ =9.11×{10}^{-31}kg×25×{10}^{14}m/s^2 \\ =22.78×{10}^{-17}N \end{gathered}$$

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