An electron having an initial horizontal velocity of magnitude$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathbf{cm}\mathbf{/}\mathbf{s}$travels into the region between two horizontal metal plates that are electrically charged. In that region, the electron travels a horizontal distance of2.00 cm and has a constant downward acceleration of magnitude$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{17}\mathbf{}}\mathbf{cm}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$due to the charged plates. Find (a)the time the electron takes to travel the 2.00 cm, (b) the vertical distance it travels during that time, and the magnitudes of its (c) horizontal and (d) vertical velocity components as it emerges from the region.

It is given that, the horizontal component of velocity is $1.0\times {10}^{17}\mathrm{cm}/{\mathrm{s}}^2$. And vertical component of velocity is zero. Horizontal distance is 2.00 cm.

The problem deals with the standard kinematic equations that describe the motion of an object with constant acceleration.By using simple kinematic equations and given data in the problem, time taken by electron can be calculated. Further, the vertical displacement of electron can be calculated by using second kinematic equation.

Formula:

The velocity and displacement in the kinematic equation are given as,

$$\begin{gathered} v_f=v_0+at \\ s=v_{0}t+\frac{1}{2}a{t}^2 \end{gathered}$$ (i)

(ii)

Where, s is total displacements,$v_0$is an initial velocity, t is time and a is an acceleration.

Using the standard relation between velocity and time,

$$\begin{gathered} \mathrm{time}=\frac{\mathrm{distance}}{\mathrm{velocity}} \\ \mathrm{t}=\frac{2.00}{1.00\times {10}^9} \\ =2.00\times {10}^{-9}\mathrm{s} \end{gathered}$$

Therefore, time taken by an electron to travel 2.00 cm distance is$2.00\times {10}^{-9}\mathrm{s}$

According to the equation (ii), the vertical displacement can be written as

$$\begin{gathered} y=v_{0}t-\frac{1}{2}a{t}^2 \\ y=0-\frac{1}{2}\left(1.00\times {10}^{17}\right)\left(2.00\times {10}^{-9}\right) \\ y=-0.20\mathrm{cm} \\ \left|\mathrm{y}\right|=0.20\mathrm{cm} \end{gathered}$$

Therefore, vertical displacement is 0.20 cm downward.

Here,

$v_x=v_0=1.00\times {10}^9\mathrm{cm}/\mathrm{s}$

Therefore, the horizontal component of velocity is$1.00\times {10}^9\mathrm{cm}/\mathrm{s}$

For the vertical component of the velocity the equation (i) can be written as,

$$\begin{gathered} v_y=0+a_{y}t \\ {v}_y=\left(1.00\times {10}^{17}\right)\left(2.00\times {10}^{-19}\right) \\ {v}_y=2.00\times {10}^8\mathrm{cm}/\mathrm{s} \end{gathered}$$

Therefore, the vertical component of velocity is$2.00\times {10}^8\mathrm{cm}/\mathrm{s}$