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)
Then Vf^2=Vi^2 + 2ad
Vf= sqrt ((2.5E4)^2+ 2*.0015*(3.1E3*1.6E-19)/(1.67262E-27))

I don't get your answer.

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 ^^

No problem! Mistakes happen to the best of us. I'm glad I could help clarify things for you. Just remember, always double-check your units and make sure you're using the correct conversions. Happy studying!

In problem #2, to find the magnitude of the electric field, you correctly used the equation V = Vo + at to find the acceleration of the proton. Then you multiplied the acceleration by the mass of the proton (1.67262 x 10^-27 kg) to find the force acting on it. Finally, you used the equation E = F/q, where q is the charge on a proton (1.6022×10^-19 C), to find the magnitude of the electric field. You obtained a value of 1.355268 x 10^-25 N/C.

For problem #3, the equation you used to find the acceleration of the proton is correct. You used the equation F = Eq = ma, where a = Eq/m. Then, to find the final speed of the proton, you used the equation Vf^2 = Vi^2 + 2ad. However, you made a mistake in the distance term. The distance should be converted to meters before plugging it into the equation. The correct distance is 0.0015 m. Once you have the correct distance, you can calculate the final speed of the proton using the equation Vf = sqrt(Vi^2 + 2ad).

In problem number 2, you correctly used the equation V = Vo + at to find the acceleration of the proton. Then, you multiplied the acceleration by the mass of the proton to find the force acting on it. However, instead of using the charge of an electron in the equation E = F/q, you should use the charge of a proton since it is a positive charge. The charge of a proton is 1.6022×10^-19 C. By substituting the values, you should get the correct answer for the magnitude of the electric field.

In problem number 3, the equation a = Eq/m is correct. It represents the acceleration of the proton due to the electric field. The charge of the proton is represented by q, which is already included in the equation. So there is no need to explicitly include the charge in the substitution.

For the equation Vf^2 = Vi^2 + 2ad, you need to use the distance in meters instead of millimeters. 1.5 mm is equal to 1.5 × 10^-3 m. By substituting the correct values and solving the equation, you should get the correct answer for the speed of the proton after it has moved a distance of 1.5 mm.