Chapter 23, Problem 1OQ

(i)

To determine

The comparison of the magnitude of electric forces exerted on the two particles.

Answer to Problem 1OQ

Option (c) They are equal.

Explanation of Solution

Write the equation for electric force.

    $F=q\times E$

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

Write the equation for the electric field on the free electron.

    $F_{\text{e}}=q_{\text{e}}\times E$                                                                                                                 (I)

Here, $F_{\text{e}}$ is the electric force on free electron particle and $q_e$ is the charge of the free electron particle.

Write the equation for the electric field on the free proton.

    $F_{\text{p}}=q_{\text{p}}\times E$                                                                                                               (II)

Conclusion:

Substitute $-q$ for $q_{\text{e}}$ in equation (I) to calculate $F_{\text{e}}$.

    $\begin{array}{c}{F}_{\text{e}}=\left(-q\right)\times E\\ =-qE\end{array}$

Substitute $q$ for $q_{\text{p}}$ in equation (II) to calculate $F_{\text{p}}$.

    $\begin{array}{c}{F}_{\text{p}}=\left(q\right)\times E\\ =qE\end{array}$

In an identical electric field, the magnitude of electric force exerted on a free electron, and on a free proton is equal.

The magnitude of electric force in free electron and free proton is equal as calculated above, therefore options (a), (b), (d) and (e) are incorrect.

Therefore, option (c) is correct.

(ii)

To determine

The comparison of magnitudes of their acceleration.

Answer to Problem 1OQ

Option (b) It is thousands of times greater for the electron.

Explanation of Solution

Write the equation for acceleration.

    $a=\frac{qE}{m}$

Here, $m$ is the particle’s mass and $a$ is the particle’s acceleration.

Particle's acceleration is inversely proportional to the mass of the particle.

    $a\propto \frac{1}{m}$

Acceleration of free electron is inversely proportional to the mass of the free electron.

    $a_{\text{e}}\propto \frac{1}{m_{\text{e}}}$                                                                                                                   (III)

Acceleration of free proton is inversely proportional to the mass of the free proton.

    $a_{\text{p}}\propto \frac{1}{m_{\text{p}}}$                                                                                                                  (IV)

Conclusion:

Substitute $9.1\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ in equation (III) to calculate $a_{\text{e}}$.

    $a_{\text{e}}\propto \frac{1}{9.1\times {10}^{-31}\text{kg}}$                                                                                                    (V)

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m_{\text{p}}$ in equation (IV) to calculate $a_{\text{p}}$.

    $a_{\text{p}}\propto \frac{1}{1.67\times {10}^{-27}\text{kg}}$                                                                                                (VI)

Divide equation (V) and (VI).

    $\begin{array}{c}\frac{a_{\text{e}}}{a_{\text{p}}}=\frac{1.67\times {10}^{-27}\text{kg}}{9.1\times {10}^{-31}\text{kg}}\\ \frac{a_{\text{e}}}{a_{\text{p}}}=1835.17\\ a_{\text{e}}=1835.17a_{\text{p}}\end{array}$                                                                                               (VII)

Mass of free proton is thousand times greater than the mass of the free electron, so the acceleration of free electron will be thousand times greater than the acceleration for free proton, as shown in equation (V), (VI) and (VII).

The acceleration of free electron will be a thousand times greater than the acceleration for free proton, therefore options (a), (c), (d) and (e) are incorrect.

Therefore, option (b) is correct.