Q50P - Fundamentals Of Physics

Question

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 by $\overrightarrow{F}=\left(120N/C\right)\hat{j}$ .In unit-vector notation, what are (a) the electron’s acceleration in that field and (b) the electron’s velocity when its x coordinate has changed by 2.0 cm?

Step-by-Step Solution

Answer

• a) The acceleration of the electron in that field is $-2.1\times {10}^{13}\frac{m}{s^2}\hat{j}$
• b) The velocity of the electron in that field is  $\left(1.5\times {10}^5 m/s\right)\hat{i}-\left(2.8\times {10}^{6}m/s\right)\hat{j}$

Step 1: The given data

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

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

Step 2: Understanding the concept of electric field

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)

Step 3: a) Calculation of the acceleration of the electron

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 $-\left(2.1\times {10}^{13}\frac{m}{s^2}\right)\hat{j}$

Step 4: b) Calculation of the velocity of the electron

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 $\left(1.5\times {10}^{5}m/s\right)\hat{i}-\left(2.8\times {10}^{6}m/s\right)\hat{j})$