Question

Chapter 22, Problem 055 A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, t.6 cm away·in a time 9.1 x 101 s (a) what is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field? (a) Number (b) Number Click if you would like to Show Work for this question: Units Units Open Show Work Question Attempts: 0 of 15 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

Only one question can be posted as per HOMEWORKLIB RULES

55)

a)

Since

$\Delta x=v_{avg}=\frac{vt}{2}=>v=\frac{2\Delta x}{t}$

v=2*0.016/(9.1*10^-8)

v=3.516*10⁵ m/s

b)

Since electric field is given by

E=F/q=ma/e

$\Delta x=\frac{1}{2}at^{2}=>a=\frac{2\Delta x}{t^{2}}$

$=>E=\frac{2m\Delta x}{et^{2}}=\frac{2\times 9.11\times 10^{-31}\times 0.016}{1.6\times 10^{-19}\times (9.1\times 10^{-8})^{2}}=22 N/C$

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66

Due to symmetry y-components cancel out each other ,the net electric field is due to x-component only

E_net=2E_x=2ECos60 =2(K|e|/d²)Cos60

E_net=2*[(9*10⁹)(1.6*10^-19)/(2.3*10^-6)²]*Cos60

E_net=272.2 N/C