An electron is projected with an initial speed${\mathit{v}}_{\mathbf{0}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s}$into the uniform field between two parallel plates (Fig. E21.29). Assume that the field between the plates is uniform and directed vertically downward and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. (a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field. (b) Suppose that the electron in Fig. E21.29 is replaced by a proton with the same initial speed${\mathit{v}}_{\mathbf{0}}$. Would the proton hit one of the plates? If not, what would be the magnitude and direction of its vertical displacement as it exits the region between the plates? (c) Compare the paths travelled by the electron and the proton, and explain the differences. (d) Discuss whether it is reasonable to ignore the effects of gravity for each particle.

$\mathrm{\Delta x}=v_{0x}t+\frac{1}{2}{a}_x{t}^2$

As horizontal acceleration=0

Therefore, time travel is:

$$\begin{gathered} t=\frac{∆x}{v_{0x}} \\ =\frac{0.02}{1.6\times {10}^6}=1.25\times {10}^{-8}s \end{gathered}$$

For the vertical acceleration;

The electric force is $\sum _{}{F}_y=m_{ay}$for vertical displacement is

$\mathrm{\Delta }y=v_{0y}t+\frac{1}{2}{a}_y{t}^2$

For vertical acceleration;

$$\begin{gathered} a_y=\frac{2\mathrm{\Delta }y}{t^2} \\ =\frac{2\times 1\times {10}^{-2}/2}{{\left(1.25\times {10}^{-8}\right)}^2}=6.4\times {10}^{13}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, for electric field is

$$\begin{gathered} E=\frac{F_0}{q_0} \\ =\frac{F_y}{e}=\frac{m{a}_y}{e} \\ =\frac{9.1\times {10}^{-31}\times 6.4\times {10}^{13}}{1.602\times {10}^{-19}} \\ =363.945\mathrm{N}/\mathrm{C} \end{gathered}$$

Therefore, the magnitude of the electric field is 363.945N/C

$\sum F_y=m{a}_y$

Thus, the electric field is

$E=\frac{F_0}{q_0}=\frac{m_{ay}}{e}$

For vertical acceleration

$$\begin{gathered} a_y=\frac{eM}{m} \\ =\frac{1.602\times {10}^{-19}\times 363.945}{1.673\times {10}^{-27}} \\ =3.485\times {10}^{10}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

For vertical displacement

$$\begin{gathered} \mathrm{\Delta }y=v_{0y}t+\frac{1}{2}{a}_y{t}^2 \\ =0+\frac{1}{2}{a}_y{t}^2 \\ =-\frac{1}{2}\times 3.48\times {10}^{10}\times {\left(1.25\times {10}^{-8}\right)}^2 \\ =-273\times {10}^{-6}\mathrm{m} \end{gathered}$$

Hence, vertical displacement of the proton is $2.723\times {10}^{-6}m$in the downward direction.

1. In an electron, the electron deflects upwards because the electric field is directed vertically upwards. And in a proton, the proton deflects downwards because the electric field is directed vertically downwards because the proton is heavier than the electron.
2. As acceleration due to electric field is much greater than the acceleration due to gravity, therefore, the influence of gravity can be minimal