Question

A uniform electric field exists in the region between two oppositely charged parallel plates 1.56 cm apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval 1.54×10^−6 s.

Part A Find the magnitude of the electric field. Use 1.60x10-19 C for the magnitude of the charge on an electron and 1.67x10-27 kg for the mass of a proton View Available Hint(s) N/C Submit PartB Find the speed of the proton at the moment t strikes the negatively charged plate View Available Hint(s) m/s Submit Provide Feedback

Answer 1

d = 0.0156meters

= ½at²

-----> 2d/t² = a = 0.0312/(1.54e-6)² = a = F/m = qE/m

m*a/q = E = [1.67e-27*0.0312/(1.54e-6)²]/1.60e-19 = 137.312V/m

E = 137.312N/C

The voltage = Ed = 137.312*0.0156 = 2.142V = 2.142J/C

2.142*1.60e-19C = 3.427e-19 J = ½*1.67e-27*V²
V² = 410419161.7

V = 20258m/s or about 20.258km/s