An electron (mass = 9.11x10 ¹ kg; charge = 1.602x10- C) is placed at the negative terminal of a parallel plate capacitor with a separation of 0.017 m. Once activated, an electric field of 1,155 N C-1 accelerates the electron towards the positive terminal. Find the magnitude of the kinetic energy of the particle when it hits the positive plate. Give your answer in electron volts (eV) noting that 1 eV = 1.602x10-1⁹ J.
An electron (mass = 9.11x10 ¹ kg; charge = 1.602x10- C) is placed at the negative terminal of a parallel plate capacitor with a separation of 0.017 m. Once activated, an electric field of 1,155 N C-1 accelerates the electron towards the positive terminal. Find the magnitude of the kinetic energy of the particle when it hits the positive plate. Give your answer in electron volts (eV) noting that 1 eV = 1.602x10-1⁹ J.
Related questions
Question
Needs Complete typed solution with 100 % accuracy.
Transcribed Image Text:An electron (mass = 9.11x10-³¹ kg; charge = 1.602x10-19 C) is placed at the negative
terminal of a parallel plate capacitor with a separation of 0.017 m. Once activated, an
electric field of 1,155 N C-1 accelerates the electron towards the positive terminal. Find the
magnitude of the kinetic energy of the particle when it hits the positive plate.
Give your answer in electron volts (eV) noting that 1 eV = 1.602x10-1⁹ J.
d
+
+
+
+
+
e
+
+
+
F=Eq
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 5 images
