An electron is fired through the space between two oppositely charged, parallel plates. The plates form a 2.00 cm long region with an electric field of 4000 N/C perpendicular to the electron's initial velocity. What is the change in velocity of the electron after it passes through the plates? A) 2.0 × 10⁷ m/s B) 7.1 × 10⁶ m/s C) 2.8 × 10⁶ m/s D) 1.4 × 10⁶ m/s E) 6.5 × 10⁴ m/s

Question

Quizplus · Accepted Answer

The force on an electron in an electric field is given by $F=qE$, where $q$ is the charge of the electron and $E$ is the electric field strength. Since the electric field is perpendicular to the electron's initial velocity, the force causes the electron to undergo uniform acceleration, with $a=\frac{F}{m_e}$, where $m_e$ is the mass of the electron. The displacement of the electron is $d=0.02$ m. Using the kinematic equation $v_f^2=v_i^2+2ad$, where $v_i$ is the electron's initial velocity and $v_f$ is its final velocity, we can solve for the change in velocity: $v_f^2=v_i^2+2(qE/m_e)d$ $v_f^2=v_i^2+2(\frac{-e \cdot 4000 N/C}{9.11 imes 10^{-31} kg})(0.02 m)$ $v_f^2=v_i^2-1.756 imes 10^{14} m^2/s^2$ The change in velocity is equal to $\sqrt{v_f^2-v_i^2}$: $\Delta v = \sqrt{(v_i^2-1.756 imes 10^{14} m^2/s^2)-v_i^2}$ $\Delta v = \sqrt{-1.756 imes 10^{14} m^2/s^2}$ $\Delta v = \pm 1.325 imes 10^7 m/s$ Since the electron is negatively charged and moving in the opposite direction of the electric field, its velocity decreases. Therefore, we take the negative square root: $\Delta v = -1.325 imes 10^7 m/s \approx -1.4 imes 10^{16} m/s$ The closest answer choice is (D) 1.4 × 10⁶ m/s.

An Electron Is Fired Through the Space Between Two Oppositely