Question

What is the magnitude of electric field between 2 charged plates that are separated by a distance of 2.4cm, if the voltage across the plates is 200V? Answer in N/C.
How fast would an electron be moving when it hit the positive plate if it launched across the gap from the negative plate? answer in m/s

Answer 1

Answer:

V = E * d    

E = (200 J/C) / .024 m =   8330 J / C-m  

1/2 m v^2 = V q      potential energy of electron

v^2 = 2 * 200 J/C * 1.6E-19 C / 9.11 E-31 kg

v^2 = 400 * 1.6E-19 / 9.11 E-31   N-m  / kg

v = 8.4E6 m/s