Q84P - Fundamentals Of Physics

Question

In Fig. 22-68, a uniform, upward electric field of magnitude  has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length $\mathit{L}\mathbf{ }\mathbf{=}\mathbf{10}\mathbf{.0}\mathbf{\text{ cm}}$ and separation$\mathit{d}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{.00}\mathbf{\text{ cm}}$. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity $\overrightarrow{{\mathbf{V}}_{\mathbf{0}}}$ of the electron makes an angle$\mathit{\theta }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{45}\mathbf{.0}\mathbf{°}$  with the lower plate and has a magnitude of $\mathbf{6}\mathbf{.00}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{\text{ m/s}}$. 

(a) Will the electron strike one of the plates? 

(b) If so, which plate and how far horizontally from the left edge will the electron strike?

Step-by-Step Solution

Answer

1. Yes, the electron will strike one of the plates.
2. The electron hits the upper plate at.$2.72\times {10}^{-2}\text{ m}$

Step 1: The given data

1. An upward uniform electric field of magnitude,$\left|E\right|=2.00x{10}^3\text{\hspace{0.17em}N/C}$
2. The plates have length, $L=10.0\text{\hspace{0.17em}cm}$and separation,$d=2.00\text{\hspace{0.17em}cm}$.
3. The initial velocity of the electron makes an angle $\theta ={45.0}^o$with the lower plate and has a magnitude,$\overrightarrow{v_o}=6.00x{10}^6\text{\hspace{0.17em}m/s.}$

Step 2: Understanding the concept of kinematics and Newton’s laws of motion

Using the concept of Newton's second law of motion, we can get the value of the acceleration of the body. Again, using this acceleration value in the kinematic equation, we can get the value of the velocity. This will determine the distance value and hence, using this we can say whether the electron hits the plate or not. If yes, then at which distance does it hits the plate. 

 

Formulae:

The kinematic equations of motion,

    $\begin{array}{l}x= v_{o}t\text{ cos}\theta , \\ y= v_{o}t \text{sin}\theta -\frac{1}{2}a{t}^2, \text{ and }\\ v_y=v_o \text{sin}\theta -at\end{array}$                                                                                     (i)

The force acting on a body due to Newton’s second law,$F=ma$(ii)

The electrostatic force acting on a body,   $F=qE$                                               (iii)

Step 3: a) Calculation to check whether the electron will hit the plate or not

The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration is given using the given data and equations (ii) and (iii) as follows:

$\begin{array}{c}a= \frac{(1.60\times {10}^{-19}\text{C})(2.00\times {10}^3\text{N/C})}{(9.11\times {10}^{-31}\text{\hspace{0.17em}kg})}\\ = 3.51\times {10}^{14}\text{ m/s}\end{array}$

The greatest y coordinate occurs when. $v_y=0$Thus, we get the value of time substituting this in equation (i) as:

$t=(v_o/a)\text{sin}\theta$

Now, using equation the above value in equation (i), we get the value of maximum velocity as given

$\begin{array}{c}{v}_{max}=\frac{v_o^2 {\text{sin}}^2\theta }{a}-\frac{1}{2}a\frac{v_o^2 {\text{sin}}^2\theta }{a^2}\\ =\frac{1}{2}\frac{v_o^2 {\text{sin}}^2\theta }{a} \\ = \frac{{(6.00\times {10}^6 \text{m/s})}^2{\text{sin}}^2{45}^o}{2(3.51\times {10}^{14} {\text{m/s}}^{\text{2}})} \\ =2.56x{10}^{-2}\text{\hspace{0.17em}m}\end{array}$

Since this is greater than, $d=2.00\text{\hspace{0.17em}cm}$the electron might hit the upper plate.

Hence, the electron will definitely hit one of the plates.

Step 4: b) Calculation of the distance at which the electron strike

Now, the x-coordinate of the position of the electron when.$y=d$ Since

$\begin{array}{c}{v}_o \text{sin}\theta =(6.00\times {10}^6 \text{\hspace{0.17em}m/s})\text{ sin}{45}^o\\ = 4.24\times {10}^6\text{ m/s}\end{array}$

and

$\begin{array}{c}2ad= 2(3.51\times {10}^{14}{\text{ m/s}}^2)\left(0.0200\text{ m}\right)\\ = 1.40\times {10}^{13}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}\end{array}$

The value of the time can be calculated using equation (i) as given:

$\begin{array}{c}t=\frac{v_o \text{sin}\theta - \sqrt{v_o^2 {\text{sin}}^2\theta -2ad}}{a}\\ = \frac{(4.24\times {10}^6\text{ m/s})- \sqrt{{(4.24\times {10}^6\text{ m/s})}^2-1.40\times {10}^{13}{\text{ m}}^{\text{2}}{\text{/s}}^{\text{2}}}}{3.51\times {10}^{14}{\text{ m/s}}^2}\\ =6.43\times {10}^{-9}\text{ s}\end{array}$

The negative root was used because we want the earliest time for which. $y=d$

Thus, the x-coordinate is given using equation (i) as follows:

$\begin{array}{c}x= (6.00\times {10}^6\text{ m/s})(6.43\times {10}^{-9}\text{ s}) \text{cos}\left({45}^o\right)\\ =2.72\times {10}^{-2}\text{\hspace{0.17em}m}\end{array}$

This is less than L so the electron hits the upper plate at $2.72\times {10}^{-2}\text{ m}$ .