Q6E - University Physics with Modern Physics

Question

An electron (mass = $\mathbf{9}\mathbf{.}\mathbf{11}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathbf{kg}$ ) 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 . 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 $\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{m}\mathbf{/}\mathbf{s}$by travelling 1.80 cm is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{14}}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$  

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

(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

 $m=9.11\times {10}^{-31}$ 

The initial velocity of the electron is

u = 0 m/s  

The final velocity of the electron is

$v=3\times {10}^{6 }\mathrm{m}/\mathrm{s}$  

The distance traveled by the electron is

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

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

    

${\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{u}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{aS}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

 

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

    

$\mathbf{v}\mathbf{=}\mathbf{u}\mathbf{+}\mathbf{at}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

 

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

F = ma          ........(3)   

Step 3: Acceleration of the electron

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

 $$\begin{gathered} \mathrm{a}=\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{S}} \\ =\frac{{\left(3\times {10}^6 \mathrm{m}/\mathrm{s}\right)}^2 -0^2}{2\times 0.018 \mathrm{m}} \\ =2.5\times {10}^{14} \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

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

Step 4: Time of motion of the electron

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

 

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

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

Step 5: Force the electron

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

$$\begin{gathered} F=ma \\ =9.11\times {10}^{-31}\mathrm{kg}\times 2.5\times {10}^{14} \mathrm{m}/{\mathrm{s}}^2 \\ =22.78\times {10}^{-17}\mathrm{N} \end{gathered}$$ 

 

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