Chapter 16, Problem 69E

(a)

To determine

The magnitude and direction of acceleration of the electron.

Answer to Problem 69E

The magnitude of acceleration is $\underset{\_}{1.76\times {10}^{14}\text{m}/{\text{s}}^2}$ and the direction is towards right.

Explanation of Solution

Write the expression for the force and electric field.

  $F=qE$        (I)

Here, $F$ is the electric force, $q$ is the charge, and $E$ is the electric field.

Write the expression for the net force acting on an object.

$F=ma$        (II)

Here, $m$ is the mass of the object and $a$ is the acceleration of the object.

Compare equations (I) and (II) to get the expression for the acceleration.

$\begin{array}{c}qE=ma\\ a=\frac{qE}{m}\end{array}$

Conclusion:

Substitute $1.6\times {10}^{-19}\text{C}$ for $q$, ${10}^3\text{N}/\text{C}$ for $E$, and $9.1\times {10}^{-31}\text{kg}$ for $m$.

  $\begin{array}{c}a=\frac{qE}{m}\\ =\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left({10}^3\text{N}/\text{C}\right)}{\left(9.1\times {10}^{-31}\text{kg}\right)}\\ =1.76\times {10}^{14}\text{m}/{\text{s}}^2\end{array}$

The direction of acceleration of the electron is same as the direction of velocity that is towards right.

Therefore, the magnitude of acceleration is $\underset{\_}{1.76\times {10}^{14}\text{m}/{\text{s}}^2}$ and the direction is towards right.

(b)

To determine

The time by the electron inside the field.

Answer to Problem 69E

The time taken by the electron inside the field is $\underset{\_}{1.0\times {10}^{-8}\text{s}}$.

Explanation of Solution

Given that the charge $-{10}^{-6}\text{C}$ is placed at the origin.

Write the expression for the velocity of electron.

  $v_0=\frac{d}{t}$        (III)

Here, $v_0$ is the velocity of electron, distance is $d$, and the time is $t$.

Write the expression for $t$ from equation (III).

$t=\frac{d}{v_0}$

Conclusion:

Substitute $0.2\text{m}$ for $d$ and $2.0\times {10}^7\text{m}/\text{s}$ for $v_0$.

  $\begin{array}{c}t=\frac{0.2\text{m}}{2.0\times {10}^7\text{m}/\text{s}}\\ =1.0\times {10}^{-8}\text{s}\end{array}$

Therefore, the time taken by the electron inside the field is $\underset{\_}{1.0\times {10}^{-8}\text{s}}$.

(c)

To determine

The vertical deflected distance as the electron leaves the field.

Answer to Problem 69E

The vertical deflected distance as the electron leaves the field is $\underset{\_}{8.8\times {10}^{-3}\text{m}}$.

Explanation of Solution

Write the equation of kinematics expression for the vertical distance deflected by the electron.

$d_y=u_{y}t+\frac{1}{2}a{t}^2$

Here, the vertical deflected distance is $d_y$ and initial vertical velocity is $u_y$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $u_y$, $1.0\times {10}^{-8}\text{s}$ for $t$, and $1.76\times {10}^{14}\text{m}/\text{s}$ for $a$.

  $\begin{array}{c}{d}_y=\left(0\text{m}/\text{s}\right)\left(1.0\times {10}^{-8}\text{s}\right)+\frac{1}{2}\left(1.76\times {10}^{14}\text{m}/{\text{s}}^2\right){\left(1.0\times {10}^{-8}\text{s}\right)}^2\\ =8.8\times {10}^{-3}\text{m}\end{array}$

Therefore, the vertical deflected distance as the electron leaves the field is $\underset{\_}{8.8\times {10}^{-3}\text{m}}$.

(d)

To determine

The angle between the actual velocity of electron and the velocity as it leaves the field.

Answer to Problem 69E

The angle between the actual velocity of electron and the velocity as it leaves the field is $5°$.

Explanation of Solution

Write the equation of kinematics expression for the vertical velocity of the electron.

$v_y=u_y+at$

Here, the vertical velocity is $v_y$.

Substitute $0\text{m}/\text{s}$ for $u_y$, $1.0\times {10}^{-8}\text{s}$ for $t$, and $1.76\times {10}^{14}\text{m}/\text{s}$ for $a$.

$\begin{array}{c}{v}_y=u_y+at\\ =\left(0\text{m}/\text{s}\right)+\left(1.76\times {10}^{14}\text{m}/\text{s}\right)\left(1.0\times {10}^{-8}\text{s}\right)\\ =1.76\times {10}^6\text{m}/\text{s}\end{array}$

Write the expression for angle between the actual velocity of electron and the velocity as it leaves the field.

$\tan \theta =\frac{v_y}{v_0}$

Write the above expression for $\theta$.

$\theta ={\tan }^{-1}\left(\frac{v_y}{v_0}\right)$

Conclusion:

Substitute $1.76\times {10}^6\text{m}/\text{s}$ for $v_y$ and $2.0\times {10}^7\text{m}/\text{s}$ for $v_0$.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{1.76\times {10}^6\text{m}/\text{s}}{2.0\times {10}^7\text{m}/\text{s}}\right)\\ =5°\end{array}$

Therefore, the angle between the actual velocity of electron and the velocity as it leaves the field is $5°$.