What is the magnitude of the acceleration of an electron when in an electric field of magnitude $7.0 \times 10^{4}$ $\mathrm{N} / \mathrm{C}$ ?

Question

Quizplus · Accepted Answer

The acceleration $a$ of an electron in an electric field $E$ can be calculated using the formula $a = \frac{F}{m}$, where $F = qE$ is the force on the electron, $q$ is the charge of the electron ($1.6 \times 10^{-19}$ C), and $m$ is the mass of the electron ($9.11 \times 10^{-31}$ kg). Substituting the given values, $a = \frac{(1.6 \times 10^{-19} \, \mathrm{C})(7.0 \times 10^{4} \, \mathrm{N/C})}{9.11 \times 10^{-31} \, \mathrm{kg}} \approx 1.2 \times 10^{16} \, \mathrm{m/s}^2$.

What Is the Magnitude of the Acceleration of an Electron 7.0×1047.0 \times 10^{4}7.0×104

What Is the Magnitude of the Acceleration of an Electron $7.0 \times 10^{4}$