# 2) A proton is moving parallel to a uniform electric field

2) A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 8.6 x 10^-23 kg·m/s from 2.4 x 10^-23 kg·m/s in a time of 5.7 x 10^-6 s. What is the magnitude of the electric field?

3) A uniform electric field has a magnitude of 3.1 x 10^3 N/C. In a vacuum, a proton begins with a speed of 2.5 x 10^4 m/s and moves in the direction of this field. Find the speed of the proton after it has moved a distance of 1.5 mm.

Here is what i did, but its not coming out correct

#2
i used the V= Vo +at to find the acceleration of the proton, then multiply that by the mass of proton (1.67262 x 10^-27 kg) to get the Force, than used the equation E = F/q where q is the charge on an electron and since its a proton its positive with the unit (1.6022×10-19 C) and i get 1.355268 x 10^-25

#3, i used the equation F = Eq = ma to find the acceleration of the proton, then used it in the equation V^2 = Vo^2 + 2ad and i got 25017.8105

2) examine the math:
Vf-Vo= at
MVf-MVo= ma*t
(MomentumF-momentumI)/t= force

(MomentumF-MomentumF/qt= E

E= (8.6E-23 – 2.4E-26)/((1.6E-19)(5.7E-6))
or about 94 volts/meter. check that.

3) a= Eq/m= (3.1E3*1.6E-19)/(1.67262E-27)
Vf= sqrt ((2.5E4)^2+ 2*.0015*(3.1E3*1.6E-19)/(1.67262E-27))

thanks i get #2 now, but in #3, a= Eq/m= (3.1E3*1.6E-19)/(1.67262E-27)
, where did the q go?

thanks i get #2 now, but in #3, a= Eq/m= (3.1E3*1.6E-19)/(1.67262E-27)
, where did the q go?

#3, a= Eq/m= (3.1E3*1.6E-19)/(1.67262E-27)
, where did the q go?

check to make certain I typed all this correctly.

ahh thank you very much, i get it now, for #3 I mistaken 1.5mm as 1.5E-6 for some reason i though it was micrometers. thank you for your help, i get it now ^^