Chapter 16, Problem 58P

(a)

To determine

The direction of the orientation of the field.

Answer to Problem 58P

The field is oriented vertically downward.

Explanation of Solution

Electric field is directed from positive charge (higher potential) to negative (lower potential). Thus, positively charged particle will move in the direction of the electric field and the negatively charged particle moves opposite direction of the electric field.

Since charge of the electron is negative, it passes through the parallel plates so the field must be oriented vertically downward direction.

Conclusion:

Therefore, the direction of the field is acting vertically downward.

(b)

To determine

The strength of the field between the parallel plates.

Answer to Problem 58P

The strength of the field between the parallel plates is $2600\text{N/C}$.

Explanation of Solution

Write the equation for the deflection time of the electron.

  $\Delta t=\frac{\Delta x}{v_i}$                                                                                          (I)

Here, $\Delta t$ is the deflection time, $\Delta x$ is the deflection distance of the electron, and $v_i$ is the initial velocity of the electron.

Write the equation for the deflection of the electron between the parallel plates.

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

Here, $d$ is the separation distance between the plates and $a_y$ is the acceleration of the electron moves vertically downward.

Rearrange the equation (II) for $a_y$.

  $a_y=\frac{2d}{\left(\Delta t^2\right)}$                                                                                   (III)

Write the expression for the electric force.

  $F_{\text{q}}=eE$  (IV)

Here, $F_{\text{q}}$ is the force due to electric field, $e$ is the charge of an electron, and $E$ is the electric field.

Write the equation for the Newton’s second law.

  $F=m{a}_y$       (V)

Here, $F$ is the force and $m$ is the mass of the electron.

Conclusion:

Substitute the equation (I) in equation (III).

  $\begin{array}{c}{a}_y=\frac{2d}{\left({\frac{\Delta x}{v_i^2}}^2\right)}\\ =\frac{2d{v}_i^2}{{\left(\Delta x\right)}^2}\end{array}$  (VI)

Compare the equation (IV) and (V) and rearrange for $E$.

  $\begin{array}{c}m{a}_y=eE\\ E=\frac{m{a}_y}{e}\end{array}$

Substitute the equation (VI) in the above equation.

  $\begin{array}{c}E=\frac{m\left(\frac{2d{v}_i^2}{\Delta x^2}\right)}{e}\\ =\frac{m}{e}\left[\frac{2d{v}_i^2}{\Delta x^2}\right]\end{array}$

Substitute $9.109\times {10}^{-31}\text{kg}$ for $m$ , $1.602\times {10}^{-19}\text{C}$ for $e$, $2.0\text{mm}$ for $d$, $8.4\times {10}^6\text{m/s}$ for $v_i$(Refer Problem 57), and $2.50\text{cm}$ for $\Delta x$ in above equation to solve for $E$.

  $\begin{array}{c}E=\frac{\left(9.109\times {10}^{-31}\text{kg}\right)}{\left(1.602\times {10}^{-19}\text{C}\right)}\left[\frac{2\left(2.0\text{mm}\right)\left(\frac{0.001\text{m}}{1\text{mm}}\right){\left(8.4\times {10}^6\text{m/s}\right)}^2}{{\left[\left(2.50\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right]}^2}\right]\\ =2565.5\text{N/C}\\ =2600\text{N/C}\end{array}$

Therefore, the strength of the field between the parallel plates is $2600\text{N/C}$.

(c)

To determine

How much less will the electrons be deflected.

Answer to Problem 58P

The deflection distance of the electron is $4.3\times {10}^{-17}\text{m}$.

Explanation of Solution

Write the equation for the deflection of the electrons due to gravitational force.

  $d=\frac{1}{2}g{\left(\Delta t\right)}^2$                                                                         (VII)

Here, $g$ is the gravitational acceleration.

Substitute equation (I) in the equation (VII).

  $d=\frac{1}{2}g{\left(\frac{\Delta x}{v_i}\right)}^2$

Conclusion:

Substitute $9.80{\text{m/s}}^2$ for $g$, $8.4\times {10}^6\text{m/s}$ for $v_i$(Refer problem 57), and $2.50\text{cm}$ for $\Delta x$ in above equation to solve for $d$.

  $\begin{array}{c}d=\frac{1}{2}\left(9.80{\text{m/s}}^2\right){\left(\frac{2.50\text{cm}\times \frac{0.01\text{m}}{1\text{cm}}}{8.4\times {10}^6\text{m/s}}\right)}^2\\ =4.3\times {10}^{-17}\text{m}\end{array}$

Therefore, the deflection distance of the electron is $4.3\times {10}^{-17}\text{m}$.