Chapter 17, Problem 109P

(a)

To determine

What is the direction and magnitude of the change in velocity of electrons between plates.

Answer to Problem 109P

The magnitude of the change in velocity is $\underset{\_}{7.0\times {10}^6\text{m}/\text{s}}$ and is directed upward.

Explanation of Solution

Write newton’s equation to find the force acting on the electrons.

    $F_y=m{a}_y$                                                                                                              (I)

Here, $F_y$ is the force, $m$ is the mass and $a_y$ is the acceleration

Write the equation to find the force due to electric field.

    $F_y=eE$                                                                                                                   (II0

Here, $e$ is charge of electron and $E$ is the electric field

Equate equations (I) and (II) and get an equation for $a_y$.

    $\begin{array}{c}eE=m{a}_y\\ a_y=\frac{eE}{m}\end{array}$                                                                                                                 (III)

Write the equation to find the time of travel of charge.

    $\Delta t=\frac{\Delta x}{v_x}$                                                                                                             (IV)

Here, $\Delta t$ is the time of travel, $\Delta x$ is change in distance and $v_x$ is the velocity

Write the equation to find the change in velocity.

  $\begin{array}{c}\Delta v_y=a_y\Delta t\\ =\left(\frac{eE}{m}\right)\left(\frac{\Delta x}{v_x}\right)\end{array}$                                                                                                  (V)

Conclusion

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$ , $2.0\times {10}^4\text{N}/\text{C}$ for $E$ , $6.0\text{cm}$ for $\Delta x$ , $9.109\times {10}^{-31}\text{kg}$ for $m$ and $3.0\times {10}^7\text{m}/\text{s}$ for $v_x$ in equation (V) to get $\Delta v_y$

    $\begin{array}{c}\Delta v_y=\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(2.0\times {10}^4\text{N}/\text{C}\right)\left(6.0\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}{\left(9.109\times {10}^{-31}\text{kg}\right)\left(3.0\times {10}^7\text{m}/\text{s}\right)}\\ =7.0\times {10}^6\text{m}/\text{s}\end{array}$

Therefore, the magnitude of the change in velocity is $\underset{\_}{7.0\times {10}^6\text{m}/\text{s}}$ and is directed upward. Since the change is positive it is upward in direction.

(b)

To determine

How far are the electrons deflected in the $\pm y$ direction between the plates.

Answer to Problem 109P

Electrons get deflected to a distance of $\underset{\_}{7.0\text{mm}}$.

Explanation of Solution

Write the equation to find the displacement of electrons.

    $\Delta y=\frac{1}{2}{a}_y{\left(\Delta t\right)}^2$                                                                                                    (VI)

Substitute equation (III) and (IV) in equation (VI) to get $\Delta y$

    $\Delta y=\frac{1}{2}\left(\frac{eE}{m}\right){\left(\frac{\Delta x}{v_x}\right)}^2$                                                                                                (VII)

Conclusion:

Substitute $1.602\times {10}^{-19}\text{C}$ for $e$ , $2.0\times {10}^4\text{N}/\text{C}$ for $E$ , $6.0\text{cm}$ for $\Delta x$ , $9.109\times {10}^{-31}\text{kg}$ for $m$ and $3.0\times {10}^7\text{m}/\text{s}$ for $v_x$ in equation (VII) to get $\Delta y$

    $\begin{array}{c}\Delta y=\frac{\left(1.602\times {10}^{-19}\text{C}\right)\left(2.0\times {10}^4\text{N}/\text{C}\right)\left(6.0\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}{2\left(9.109\times {10}^{-31}\text{kg}\right)\left(3.0\times {10}^7\text{m}/\text{s}\right)}\\ =7.0\text{mm}\end{array}$

Therefore, electrons get deflected to a distance of $\underset{\_}{7.0\text{mm}}$.