Question: At some instant the velocity components of an electron moving between two charged parallel plates are$V_X=1.5\times {10}^{5}m/s$ and$V_y=3.0\times {10}^{3}m/s$. Suppose the electric field between the plates is uniform and given byrole="math" localid="1662552218156" $\overrightarrow{F}=\left(120N/C\right)\hat{j}$.In unit-vector notation, what are (a) the electron’s acceleration in that field and (b) the electron’s velocity when itsxcoordinate has changed by 2.0 cm?

The velocity components of the electron,and$V_X=1.5\times {10}^{4}m/sand{V}_y=3.0\times {10}^{3}m/s$

The force between the plates,$\overrightarrow{E}=(120N/C)\hat{j}$

We assume there are no forces or force components along the x-direction. We combine the equation for the magnitude of the electrostatic force on a point charge of magnitude q with Newton’s second law, then use the concept of projectile motion to determine the time t taken by the particle and the final velocity (with –g replaced by the y-component of the acceleration of this problem). For these purposes, the velocity components given in the problem statement are re-labelled as v0x and v0y, respectively.

Formula:

The force on a particle due to electric field, F =qE (i)

The force due to Newton’s second law, F =ma (ii)

The distance travelled by the particle using projectile motion,

$x-x_o=(v_o\cos {\theta }_o)t$ (iii)

Using the given data and equation (i) in equation (ii), we can get the acceleration of the particle as:
$$\begin{gathered} \overrightarrow{a}=-\left(\frac{1.60\times {10}^{-19}C}{9.11\times {10}^{-31}kg}\right)\left(120\frac{N}{C}\right)\overbrace{j} \\ =-\left(2.1\times {10}^{13}\frac{m}{s^2}\right)\hat{j} \end{gathered}$$

Hence, the value of the magnitude of the acceleration is$-\left(2.1\times {10}^{13}\frac{m}{s^2}\right)\hat{j}$

Since,$V_x=V_{0x}$in this problem (that is$a_x=0$,), we obtain the time taken by the electron using equation (iii) is given as:

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{t}=\frac{△\mathrm{x}}{{\mathrm{V}}_{0\mathrm{X}}} \\ =\frac{0.020\mathrm{m}}{1.5\times {10}^5\mathrm{m}/\mathrm{s}} \\ =1.3\times {10}^{-7}\mathrm{s} \end{gathered}$$

Thus, the y-component of the velocity of the electron is given using equation (iv) as:

$$\begin{gathered} V_y=V_{0y}+a_{y}t \\ =3.0\times \frac{{10}^{3}m}{s}+\left(-2.1\times {10}^{13}\frac{m}{s^2}\right)\left(1.3\times {10}^{-7}s\right) \\ =-2.8\times {10}^{6}m/s \end{gathered}$$

Now, the final velocity of the electron is given as:
$\overrightarrow{V}=\left(1.5\times {10}^{5}m/s\right)\hat{i}-\left(2.8\times {10}^{6}m/s\right)\hat{j}$

Hence, the value of the velocity is$\left(1.5\times {10}^{5}m/s\right)\hat{i}-\left(2.8\times {10}^{6}m/s\right)\hat{j})$