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Speed of a Proton in An Electric Field

  • Thread starter servb0t
  • Start date
  • #1
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Hello,

First time posting, so I apologize in advance for anything wacky or off. Basically, I'm doing an online homework assignment for my class. Haven't taken physics in three years (hooray high school!), so I'm a bit rusty.

I've laid out the math pretty much exactly how I've attempted it in my notebook, as I believe that my error(s) are not based in the equations I've used.

If any of you wiz kids could kindly look this over and gently guide me in the right direction, I'm sure I'll be able to figure it out. Thanks in advance!


1)
A uniform electric field has a magnitude of 2.35E+3 N/C. In a vacuum, a proton begins with a speed of 2.55E+4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 2.20 mm.




2)
Force=(Proton Charge)*(Efield)

F=ma

(Vf)^2=(Vi)^2 + 2ax




3)
Force=(Proton Charge)*(Efield)
F= (+1.6e-19 C)*(2.53e3 N/C)= 4.048e-16 N

F=(mass of proton)*(acceleration of proton)
(4.048e-16 N) = (1.673e-27kg)*(a)
(a) = (4.048e-16 N) / (1.673e-27 kg)
a = 2.424e11 m/s/s

(Vf)^2 = (Vi)^2 + 2ax
(Vf)^2 = (2.55e4 m/s)^2 + 2(2.424e11 m/s/s)(2.2e-3 m)
(Vf)^2 = (6.5e8) + 2(5.3e8)
(Vf)^2 = (6.5e8) + (1.065e9)
(Vf)^2 = 1.72e9 m/s
(Vf)= sqrt(1.72e9 m/s)
(Vf) = 41,413.16 m/s
 

Answers and Replies

  • #2
2
0
Bah, looking over what I wrote in the thread, it's very apparent that I used E= 2.53e3 N/C instead of E=2.35e3 N/C in the calculations. I revised my math and got the right answer (4.05e4 m/s.)

I'm sure I'll be here more in the future though, thanks for anyone who cared to read :)

/thread
 

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