Q6 E - University Physics with Modern Physics

Question

An electron (mass = $\mathbf{9}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathit{k}\mathit{g}$) leaves one end of a TV picture tube with zero initial speed and travels in a straight line to the accelerating grid, which is 1.80 cm away. It reaches the grid with a speed of $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s}$ . If the accelerating force is constant, compute

 

(a) the acceleration; 

 

(b) the time to reach the grid; and 

 

(c) the net force, in newtons. Ignore the gravitational force on the electron.

Step-by-Step Solution

Answer

(a) The acceleration of the electron reaching a speed of $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathit{m}\mathbf{/}\mathit{s}$  by travelling $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{c}\mathit{m}$ is  $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$

(b) The time taken by the electron to reach this speed is  $\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathit{s}$

(c) The force on the electron to produce this acceleration is $\mathbf{22}\mathbf{.}\mathbf{78}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{17}}\mathbf{N}$ .

Step 1: Given data

The mass of the electron is

$m=9.11\times {10}^{-31}kg$  

The initial velocity of the electron is

  u = 0 m/s

The final velocity of the electron is

$v=3\times {10}^{6}m/s$  

The distance traveled by the electron is

 $$\begin{gathered} s=1.8 cm \\ =1.8.\left(1 cm\times \frac{1m}{100 cm}\right) \\ =0.018 m \end{gathered}$$ 

Step 2: Equations of motion and second law of motion

The initial velocity u , final velocity v , acceleration  a and the distance traveled S  are related as

    ${\mathit{V}}^{\mathbf{2}}\mathbf{=}{\mathit{u}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{a}\mathit{s}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)$

 

The initial velocity u , final velocity v , acceleration  a and the time of travel  t are related as

   $\mathit{v}\mathbf{=}\mathit{u}\mathbf{+}\mathit{a}\mathit{t}\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(2\right)$ 

 

According to the second law of motion, the force on an object of mass  and acceleration   is

   $\mathit{F}\mathbf{=}\mathit{m}\mathit{a}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(3\right)$

Step 3: Acceleration of the electron

Let the acceleration of the electron be a . From equation (1),

$$\begin{gathered} a=\frac{v^2-u^2}{2S} \\ =\frac{{\left(3\times {10}^{6}m/s\right)}^2}{2\times 0.018 m} \\ =2.5\times {10}^{14}m/s^2 \end{gathered}$$ 

 

Thus, the acceleration is $2.5\times {10}^{14}m/s^2$ .

Step 4: Time of motion of the electron

Let the time of motion of the electron be t . From equation (2),

$$\begin{gathered} t=\frac{v-u}{a} \\ =\frac{3\times {10}^{6}m/s-0}{2.5\times {10}^{14}m/s^2} \\ =1.2\times {10}^{-8}s \end{gathered}$$ 

 

Thus, the time of motion of the electron is $1.2\times {10}^{-8}s$ .

Step 5: Force the electron

From equation (3), the force on the electron is

 $$\begin{gathered} F=ma \\ =9.11\times {10}^{-31}kg\times 25\times {10}^{14}m/s^2 \\ =22.78\times {10}^{-17}N \end{gathered}$$

 

Thus, the force on the electron is $22.78\times {10}^{-17}N$ .