- #1
Feodalherren
- 605
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Homework Statement
A proton is projected in the positive x direction into a region of uniform electric field E = (-5.80 ✕ 10^5) N/C at t = 0. The proton travels 6.50 cm as it comes to rest.
Find the initial velocity and the time it takes for the proton to stop.
Homework Equations
The Attempt at a Solution
The charge of the proton multiplied by the electric field must equal the mass of the proton times the acceleration
qE=ma
hence
a= (-5.556x10^13)
Constant accelecration, integrate with respect to time to get velocity
V= Vi - 5.556x10^13 t
X= 0 + Vi(t) - 2.2778 t^2
Since initial position = 0.
Using the position function to find the time it takes to stop:
.065m = (5.556x10^13)t^2 - (2.2778x10^13)t^2
solving for t, t= 4.8x10^-8 s
Plugging this back to the velocity equation where at t= 4.8x10^-8 s the final velocity = 0 therefore:
Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)
Clearly not right, it's faster than the SOL.
Where am I going wrong?
ps. I don't like to memorize stuff so please don't refer me to the kinematic equations. I prefer to derive them by myself.