- #1

servb0t

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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.

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

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

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