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Problem 28.2 Part A The electric field strength is 2.20x104 N/C inside a parallel-plate capacitor with a 1.20 mm spacing. An electron is released from rest at the negative plate. What is the electrons speed when it reaches the positive plate? Express your answer with the appropriate units alueUnits Submit My Answers Give UpThe electric field strength is 2.20×104 N/C inside a parallel-plate capacitor with a 1.20 mm spacing. An electron is released from rest at the negative plate.

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Answer #1

F = qE = 1.6*(10^-19) * 2.20×10^4 = 3.52*(10^-15) N

F = ma
or, a = F/m = [3.52 *(10^-15)] / [9.1 × (10^-31)] = 3.868 * (10^15) m/s^2

v^2= u^2 + 2ax = 0 + 2*[3.868 * (10^15)]*[1.2*(10^-3)] = 9.28* (10^12)
or, v = sqrt (9.28* (10^12) ) = 3.046 * (10^6) m/s

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