.\'ll Sprint 8:17 PM a session.masteringphysics.com C Ch 21 t A Proten between O
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.'ll Sprint 8:17 PM a session.masteringphysics.com C Ch 21 t A Proten between Oppositely Charged Plates A uniform electric field exists in the region between two oppositely charged parallel plates 1.56 em 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 1.41×10-6 s . Part A Find the magnitude of the electric field Use 1.60*10-19 C for the magnitude of the charge on an electron and 1.67 10-27 kg for the mass of a proton View Available Hint(s) N/C Part B Find the speed of the proton at the moment it strikes the negatively charged plate. View Available Hint(s) Provide FeedbackExplanation / Answer
From Second equation of kinematics: S= ut+(1/2)at2
As electron is released from rest, so u= 0
a= 2S/t2
S = 0.0156 meter, t= 1.41x10-6 second
(a) Velocity V= u+at = at= (2S/t2)*t = 2S/t = 2x0.0156/1.41x10-6 = 22127.66 m/s
(b) Force on electron in electric filed = qE
From newton equation ma= qE
E= ma/q
m= mass of proton = 1.67x10-27kg, q= charge on electron= 1.6x10-19 C
E= 2mS/qt2 = 2x 1.67x10-27x 0.0156/1.6x10-19x(1.41x10-6)2 = 163.80 N/C
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