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 $500 \, \text{N/C}$. An electron (charge $e = 1.6 \times 10^{-19} \, \text{C}$, mass $m_e = 9.1 \times 10^{-31} \, \text{kg}$) 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 $F_e = qE$. $$F_e = (1.6 \times 10^{-19} \, \text{C}) \times (500 \, \text{N/C})$$ $$F_e = 8.0 \times 10^{-17} \, \text{N}$$ Step 2: Calculate the acceleration of the electron. According to Newton's second law, $F_e = m_e a$, where $a$ is the acceleration. $$a = \frac{F_e}{m_e}$$ $$a = \frac{8.0 \times 10^{-17} \, \text{N}}{9.1 \times 10^{-31} \, \text{kg}}$$ $$a \approx 8.79 \times 10^{13} \, \text{m/s}^2$$ Step 3: Calculate the time taken to travel the distance. The distance $d$ between the plates is $15 \, \text{cm} = 0.15 \, \text{m}$. Since the electron starts from rest, we can use the kinematic equation $d = v_0 t + \frac{1}{2}at^2$. Here, $v_0 = 0$. $$d = \frac{1}{2}at^2$$ $$t^2 = \frac{2d}{a}$$ $$t = \sqrt{\frac{2d}{a}}$$ $$t = \sqrt{\frac{2 \times 0.15 \, \text{m}}{8.79 \times 10^{13} \, \text{m/s}^2}}$$ $$t = \sqrt{\frac{0.30}{8.79 \times 10^{13}}} \, \text{s}$$ $$t = \sqrt{3.413 \times 10^{-15}} \, \text{s}$$ $$t \approx 5.84 \times 10^{-8} \, \text{s}$$ The time taken is $\boxed{\text{5.84} \times 10^{-8} \, \text{s}}$. 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 $v_0 = 0$: $$v = at$$ $$v = (8.79 \times 10^{13} \, \text{m/s}^2) \times (5.84 \times 10^{-8} \, \text{s})$$ $$v \approx 5.13 \times 10^6 \, \text{m/s}$$ The final speed of the electron is $\boxed{\text{5.13} \times 10^6 \, \text{m/s}}$. Send me the next one 📸