Calculating Proton's Motion in Uniform Electric Field - Answer Check & Help

[cowgiljl]

A proton traveling at 3E6 m/s enters a region where the electric field has a magnitude of 3E5 N/C. The electric field is uniform and slows the proton's motions.

a)calculate the distamce the proton will travel before coming to a momentary halt?

B)Calculate the decel. of the proton and the time for it to trqavel this distance

F=qE = 1.6E-19 * 3E5 = 4.8E-14 N

a = F/m 4.8E-14 / 1.67E-31 a = 2.87E13 m/s

D = Vf^2 - Vi^2 / 2A = 3E6^2 / 2 * 2.87E13 = .157 m

I am not sure how to do part B or what formula to use

thanks joe

 

[Chen]

Assuming "decel." means deceleration, you already found it in part (a). To find the time it travels before stopping, use this formula:
$$V_t = V_0 + at$$

 

[M98Ranger]

For part B, you can use the equation for deceleration, which is a = (Vf - Vi)/t, where a is the deceleration, Vf is the final velocity, Vi is the initial velocity, and t is the time.

To find the deceleration, you can use the same formula as in part A, which is a = F/m. So, a = 2.87E13 m/s.

Next, you can use the distance calculated in part A, which is 0.157 m, as the displacement (A) in the deceleration equation. The final velocity (Vf) is 0 m/s since the proton comes to a halt, and the initial velocity (Vi) is 3E6 m/s.

Plugging these values into the equation, you get: 2.87E13 = (0 - 3E6)/t

Solving for t, you get t = 1.04E-7 seconds. This is the time it takes for the proton to come to a halt.

Hope this helps!