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C Secure https://session masteringphysics.com/myct/itemview?assig University Physics ll PHYS2326: SPRING 2017 Ch 21 HW Exercise 21.25 Enhanced with Solution Exercise 21.25 Enhanced with Solution A proton is traveling horizontally to the right at You may want to review (u pages 695-699) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electron in a uniform field. O Type or speak to me

Solution

Data:
Initial speed, V_i = 4.8 x 10⁶ m/s
Charge of the proton, q = 1.6 x 10^-19 C
Mass of the proton, m = 1.67 x 10^-27 kg

Solution:
PART (a)
Distance, s = 0.034 m
Final speed, V_f = 0 m/s
Acceleration, a = F / m
= q E / m
= (1.6 x 10^-19)E / (1.67 x 10^-27)
= 9.58 x 10⁷ E
From the equations of motion,
V_f² - V_i² = 2as
0 - (4.8 x 10⁶ )² = 2(9.58 x 10⁷ E )(0.034)
So, E = - 3.537 x 10⁶ N/C
Magnitude of electric field E = 3.537 x 10⁶ N/C
Direction: Horizontally towards left

PART (b)
The angle counterclockwise from left is zero.

PART (c)
Acceleration, a = q E / m
= (1.6 x 10^-19)(-3.537 x 10⁶) / (1.67 x 10^-27)
= - 3.388 x 10¹⁴ m/s²
From the equation of motion,
V_f = V_i + at
0 = 4.8 x 10⁶ + ( - 3.388 x 10¹⁴ ) * t
Time, t = 1.42 x 10^-8 s

PART (d)
Mass of electron, m = 9.1 x 10^-31 kg
Charge of electron, q = 1.6 x 10^-19 C
Distance, s = 0.034 m
Final speed, V_f = 0 m/s
Acceleration, a = F / m
= q E / m
= (1.6 x 10^-19) E / (9.1 x 10^-31)
= (1.76 x 10¹¹) E
From the equations of motion,
V_f² - V_i² = 2as
0 - (4.8 x 10⁶)² = 2 (1.76 x 10¹¹ E)(0.034)
So, E = - 1925.13 N/C
Magnitude of electric field = 1925.13 N/C
Direction: Horizontally towards the right