Question

Part A Review A proton is released from rest in a uniform electric field of magnitude 2.18x105 N/C. Find the speed of the proton after it has traveled (a) 1.10 cm and (b) 14.0 cm. m/s Submit Request Answer Part B m/s

Answer 1

Proton is released from rest so u = 0 m/s = initial speed

E = 2.18 x 10⁵ N/C

F = E*q
F = 2.18x10⁵*(1.6x10^-19)
F = 3.488 x 10^-14 N

and we also know that

F=m*a
a = F/m
a = (3.488 x 10^-14)/(1.672 x 10^-27)
a = 2.086124 x 10¹³ m/s²

using a in

v² = u² + 2*a*s

v = (2*a*s)

v = (2*2.086124 x 10¹³*0.011)

v = 677456.54 m/s

v = 6.77 x 10⁵ m/s

(b)

same for this part s = 0.14 m

v² = u² + 2*a*s

v = (2*a*s)

v = (2*2.086124 x 10¹³*0.14)

v = 2416846.77 m/s

v = 2.41 x 10⁶ m/s