The problem describes two charged metal plates in a vacuum, 15 cm apart. The electric field between the plates is uniform and has a strength of 500 N/C. An electron (charge $e = 1.6 x

Welcome back offikwuvictoria081 — missed you this week. The problem describes two charged metal plates in a vacuum, 15 cm apart. The electric field between the plates is uniform and has a strength of 500 \, N/C. An electron (charge e = 1.6 × 10^-19 \, C, mass m_e = 9.1 × 10^-31 \, kg) is released from rest just outside the negative plate. We need to find: a) How long it will take to reach the positive plate. b) How fast the electron will be moving when it reaches the positive plate. a) Time to reach the positive plate Step 1: Calculate the electric force on the electron. The electric force F_e on a charge q in an electric field E is given by F_e = qE. F_e = (1.6 × 10^-19 \, C) × (500 \, N/C) F_e = 8.0 × 10^-17 \, N Step 2: Calculate the acceleration of the electron. According to Newton's second law, F_e = m_e a, where a is the acceleration. a = (F_e)/(m_e) a = 8.0 × 10^-17 \, N9.1 × 10^-31 \, kg a ≈ 8.79 × 10^13 \, m/s^2 Step 3: Calculate the time taken to travel the distance. The distance d between the plates is 15 \, cm = 0.15 \, m. Since the electron starts from rest, we can use the kinematic equation d = v_0 t + (1)/(2)at^2. Here, v_0 = 0. d = (1)/(2)at^2 t^2 = (2d)/(a) t = sqrt((2d)/(a)) t = sqrt(2 × 0.15 \, m)8.79 × 10^13 \, m/s^2 t = sqrt((0.30)/(8.79 × 10^13)) \, s t = sqrt(3.413 × 10^-15) \, s t ≈ 5.84 × 10^-8 \, s The time taken is 5.84 × 10^-8 \, s. b) Final speed of the electron Step 1: Use the kinematic equation to find the final velocity. We can use the equation v = v_0 + at. Since v_0 = 0: v = at v = (8.79 × 10^13 \, m/s^2) × (5.84 × 10^-8 \, s) v ≈ 5.13 × 10^6 \, m/s The final speed of the electron is 5.13 × 10^6 \, m/s. Send me the next one 📸