Question

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.

Short answer

a) Time taken by electron to travel 2.00 cm distance is$2.00\times {10}^{-9}\mathrm{s}$

b) Vertical displacement is$0.20\mathrm{cm}$ downward.

c) Magnitude of its horizontal velocity components is$1.00\times {10}^9\mathrm{cm}/\mathrm{s}$

d) Magnitude of vertical velocity components as it emerges from the region is$2.00\times {10}^8\mathrm{cm}/\mathrm{s}$

Step-by-step solution

Given information

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.

Determining the concept

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.

(a) determining the time to travel  horizontal distance

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$2.00\times {10}^{-9}\mathrm{s}$

(b) Determining thevertical displacement

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.

(c) Determining thehorizontal component of velocity

Here,

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

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

(d) Determining thevertical component of velocity

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$2.00\times {10}^8\mathrm{cm}/\mathrm{s}$