Chapter 19, Problem 32P

(a)

To determine

The analysis model that describes the horizontal motion of the protons above the plane.

Answer to Problem 32P

The particle under constant velocity describes the horizontal motion of the protons above the plane.

Explanation of Solution

Figure 1 represents a proton is projected with an angle of $\theta$ with initial velocity $v_i$, under an uniform electric field $\overrightarrow{\text{E}}$.

Write the expression for horizontal component of motion of the proton.

    $v_{\text{x}}=v_{\text{0}}\cos \theta$

Here, $v_{\text{x}}$ is the horizontal component, and $v_{\text{0}}$ is the initial velocity.

There is no force acting on the proton in the horizontal direction, thus the particle under constant velocity describes the horizontal motion of the protons above the plane.

Conclusion:

Therefore, the particle under constant velocity describes the horizontal motion of the protons above the plane

(b)

To determine

The analysis model that describes the vertical motion of the protons above the plane.

Answer to Problem 32P

The particle under constant acceleration describes the vertical motion of the protons above the plane.

Explanation of Solution

Write the expression for vertical component of motion of the proton.

    $v_{\text{y}}=v_{\text{0}}\sin \theta -gt$

Here, $v_y$ is the vertical component of motion, $g$ is the acceleration due to gravity, and $t$ is the time.

From the above equation it is clear that the vertical component of the velocity depends only on acceleration due to gravity. The acceleration due to gravity has a constant value throughout the motion, thus the particle under constant acceleration describes the vertical motion of the protons above the plane.

Conclusion:

Therefore, the particle under constant acceleration describes the vertical motion of the protons above the plane.

(c)

To determine

Whether the Equation 3.16 be applicable to the protons.

Answer to Problem 32P

Yes, the Equation 3.16 is applicable to the protons. The proton moves in a parabolic path.

Explanation of Solution

Given that the electric field is $\overrightarrow{\text{E}}=-720\widehat{\text{j}}\text{N}/\text{C}$.

The vertical acceleration caused by the constant electric force $\overrightarrow{\text{E}}=-720\widehat{\text{j}}\text{N}/\text{C}$, and this this electric field is in the downward direction.

This vertical acceleration makes the proton to move in a parabolic path, this is similar to a projectile in a gravitational field.

Conclusion:

Therefore, the Equation 3.16 is applicable to the protons. The proton moves in a parabolic path.

(d)

To determine

The expression for range $R$ in terms of initial velocity, electric field, mass of proton, the charge, and the angle.

Answer to Problem 32P

The expression for range $R$ in terms of initial velocity, electric field, mass of proton, the charge, and the angle is $\underset{\_}{R=\frac{m_p{v}_i^2\sin 2\theta }{eE}}$.

Explanation of Solution

Write the expression for vertical acceleration.

    $a_y=\frac{eE}{m_p}$        (I)

Here, $a_y$ is the vertical acceleration, $e$ is the charge of the proton, $E$ is the electric field, and $m_p$ is the mass of proton.

Write the expression for range from Equation 3.16.

    $R=\frac{v_i^2\sin 2\theta }{g}$        (II)

Here, $R$ is the range, $v_i$ is the initial velocity, and $g$ is the acceleration due to gravity.

Since the vertical acceleration is greater than acceleration due to gravity, consider vertical acceleration in the place of acceleration due to gravity.

Apply the above condition in equation (II)

    $\begin{array}{c}R=\frac{v_i^2\sin 2\theta }{\frac{eE}{m_p}}\\ =\frac{m_p{v}_i^2\sin 2\theta }{eE}\end{array}$        (III)

Conclusion:

Substitute $1.60\times {10}^{-19}\text{C}$ for $e$, $720\text{N}/\text{C}$ for $E$, and $1.67\times {10}^{-27}\text{kg}$ for $m_p$ in equation (I), to find $a_y$.

    $\begin{array}{c}{a}_y=\frac{\left(1.60\times {10}^{-19}\text{C}\right)\left(720\text{N}/\text{C}\right)}{\left(1.67\times {10}^{-27}\text{kg}\right)}\\ =6.90\times {10}^{10}\text{m}/{\text{s}}^2\end{array}$

Therefore, the expression for range $R$ in terms of initial velocity, electric field, mass of proton, the charge, and the angle is $\underset{\_}{R=\frac{m_p{v}_i^2\sin 2\theta }{eE}}$.

(e)

To determine

The two possible values of the angle $\theta$.

Answer to Problem 32P

The two possible values of the angle $\theta$ are $\underset{\_}{36.9°}$, and $\underset{\_}{53.1°}$.

Explanation of Solution

From subpart (d) the expression for range $R$ in terms of initial velocity, electric field, mass of proton, the charge, and the angle is given by $\underset{\_}{R=\frac{m_p{v}_i^2\sin 2\theta }{eE}}$.

Given that the range is $1.27\text{mm}$.

Conclusion:

Substitute $1.67\times {10}^{27}\text{kg}$ for $m_p$, $9.55\times {10}^3\text{m}/\text{s}$ for $v_i$, $1.60\times {10}^{-19}\text{C}$ for $e$, $1.27\text{mm}$ for $R$ and $720\text{N}/\text{C}$ for $E$ in equation (III), to find $\theta$.

    $\begin{array}{c}1.27\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}=\frac{\left(1.67\times {10}^{27}\text{kg}\right){\left(9.55\times {10}^3\text{m}/\text{s}\right)}^2\sin 2\theta }{\left(1.60\times {10}^{-19}\text{C}\right)\left(720\text{N}/\text{C}\right)}\\ 1.27\times {10}^{-3}\text{m}=\frac{\left(1.67\times {10}^{27}\text{kg}\right){\left(9.55\times {10}^3\text{m}/\text{s}\right)}^2\sin 2\theta }{\left(1.60\times {10}^{-19}\text{C}\right)\left(720\text{N}/\text{C}\right)}\\ \sin 2\theta =0.961\\ \theta =36.9°\end{array}$

The other value of $\theta$ is $90-\theta$ which is $53.1°$.

Therefore, the two possible values of the angle $\theta$ are $\underset{\_}{36.9°}$, and $\underset{\_}{53.1°}$.

(f)

To determine

The time interval during which the proton is above the plane for the two possible values of $\theta$.

Answer to Problem 32P

The time interval during which the proton is above the plane for the two possible values of $\theta$ are $\underset{\_}{166\text{ns}}$, and $\underset{\_}{221\text{ns}}$.

Explanation of Solution

Write the expression for time interval.

    $\Delta t=\frac{R}{v_i\cos \theta }$        (IV)

Here, $\Delta t$ is the time interval, $R$ is the range, and $v_i$ is the initial velocity.

Conclusion:

Substitute $1.27\times {10}^{-3}\text{m}$ for $R$, $9.55\times {10}^3\text{m}/\text{s}$ for $v_i$, and $36.9°$ for $\theta$ in equation (IV), to find $\Delta t$.

    $\begin{array}{c}\Delta t=\frac{1.27\times {10}^{-3}\text{m}}{9.55\times {10}^3\text{m}/\text{s}\times \cos 36.9°}\\ =1.66\times {10}^{-7}\text{s}\times \frac{1\text{ns}}{{10}^{-9}\text{s}}\\ =166\text{ns}\end{array}$

Substitute $1.27\times {10}^{-3}\text{m}$ for $R$, $9.55\times {10}^3\text{m}/\text{s}$ for $v_i$, and $53.1°$ for $\theta$ in equation (IV), to find $\Delta t$.

    $\begin{array}{c}\Delta t=\frac{1.27\times {10}^{-3}\text{m}}{9.55\times {10}^3\text{m}/\text{s}\times \cos 53.1°}\\ =2.21\times {10}^{-7}\text{s}\times \frac{1\text{ns}}{{10}^{-9}\text{s}}\\ =221\text{ns}\end{array}$

Therefore, the time interval during which the proton is above the plane for the two possible values of $\theta$ are $\underset{\_}{166\text{ns}}$, and $\underset{\_}{221\text{ns}}$.