To determine the time interval over which the proton comes to rest, we can use the equation for the acceleration of a charged particle in an electric field:
a = qE / m
where a is the acceleration, q is the charge of the proton, E is the electric field, and m is the mass of the proton.
First, we need to determine the acceleration of the proton. The charge of a proton is q = 1.602 x 10^-19 C, and the mass of a proton is m = 1.6726 x 10^-27 kg. Plugging these values into the equation, we get:
a = (1.602 x 10^-19 C)(-6 x 10^5 N/C) / (1.6726 x 10^-27 kg)
≈ -5.72 x 10^16 m/s^2 (negative because the acceleration is in the opposite direction of the electric field)
Next, we need to determine the initial velocity of the proton. We are given that At1 = 0, which means the proton starts from rest. Therefore, the initial velocity is v0 = 0 m/s.
We can use the kinematic equation to find the time interval, t, over which the proton comes to rest:
v = v0 + at
Since v0 = 0 m/s and a = -5.72 x 10^16 m/s^2, the equation becomes:
v = -5.72 x 10^16 m/s^2 * t
We are given that the proton travels a distance of 7.00 cm = 0.07 m. We can use another kinematic equation to relate the acceleration, initial velocity, time, and displacement:
s = v0t + 0.5at^2
Since v0 = 0 m/s and s = 0.07 m, the equation becomes:
0.07 m = 0 + 0.5(-5.72 x 10^16 m/s^2)t^2
Simplifying, we get:
0.07 m = -2.86 x 10^16 m/s^2 * t^2
Dividing both sides by -2.86 x 10^16 m/s^2, we get:
t^2 = 0.07 m / -2.86 x 10^16 m/s^2
≈ -2.44 x 10^-19 s^2
Taking the square root of both sides, we get:
t = √(-2.44 x 10^-19 s^2)
≈ ± 4.94 x 10^-10 s
Since time cannot be negative, we only consider the positive value as the time interval over which the proton comes to rest:
t ≈ 4.94 x 10^-10 s
Therefore, the time interval over which the proton comes to rest is approximately 4.94 x 10^-10 seconds.
At1=0, a proton is projected in the positive x-direction into a region of a uniform electric field of E=-6x1051. The proton travels 7.00 cm as it comes to rest. Determine The time interval over which the proton comes to rest
in the simplest steps
1 answer