Chapter 16, Problem 27P

To determine

To Find: The magnitude of acceleration experienced by an electron in an electric field.

Answer to Problem 27P

  $a=1.32\times {10}^{14}\text{m}/{\text{s}}^2$

The electron accelerates in a direction opposite to that of the applied electric field.

Explanation of Solution

Given:

Electric field, $E=750\text{N}/\text{C}$

Formula used:

Force on electron in an electric field:

  $\begin{array}{l}F=m_{e}a=eE\\ a=\frac{eE}{m_e}\end{array}$

Calculation:

Charge of electron, $e=-1.6\times {10}^{-19}\text{C}$

Mass of electron, $m_e=9.11\times {10}^{-31}\text{kg}$

Electric field applied, $E=750\text{N}/\text{C}$

The acceleration of the electron:

  $a=\frac{eE}{m_e}$

  $a=\frac{\left(-1.6\times {10}^{-19}\right)(750)}{\left(9.11\times {10}^{-31}\right)}\text{m}/{\text{s}}^2$

  $a=-1.32\times {10}^{14}\text{m}/{\text{s}}^2$

The negative sign indicates that the electron accelerates in a direction opposite to that of the applied electric field.