A proton is projected in the positive x direction into a region of uniform electric field E = (-6.80 * 10^5 N/C at t = 0. The proton travels 6.80 cm as it comes to rest.
(a) Determine the acceleration of the proton.
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To determine the acceleration of the proton, we can use the equation of motion that relates acceleration, initial velocity, final velocity, and displacement:
vf^2 = vi^2 + 2ad
In this case, the proton is initially moving in the positive x direction and is brought to rest, so its final velocity, vf, is 0 m/s. The initial velocity, vi, is unknown.
The displacement, d, is given as 6.80 cm, which can be converted to meters by dividing by 100:
d = 6.80 cm = 0.0680 m
Plugging in these values into the equation, we can solve for acceleration (a):
0 = vi^2 + 2ad
Rearranging the equation:
vi^2 = -2ad
Substituting the given value of the electric field:
vi^2 = 2(-6.80 * 10^5 N/C)(0.0680 m)
Now, we can solve for vi by taking the square root of both sides of the equation:
vi = √[2(-6.80 * 10^5 N/C)(0.0680 m)]
Calculating this expression will give us the initial velocity, vi.
Once we have calculated vi, we can use the kinematic equation to find the acceleration, a, by substituting the values of vi, vf, and d into the equation:
vf = vi + at
Since vf is 0, the equation simplifies to:
0 = vi + at
Solving for a:
a = -vi/t
In this case, t = 0, so the acceleration will be zero.