Question
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
in the simplest steps
Answers
GPT 3.5
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.
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.