A proton moves at 3.60 *10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.20* 10^3 N/C.
Find the time interval required for the proton to travel 4.00 cm horizontally.
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To find the time interval required for the proton to travel 4.00 cm horizontally, we need to use the information given:
Speed of the proton (v) = 3.60 * 10^5 m/s
Distance traveled horizontally (d) = 4.00 cm = 0.04 m
Electric field strength (E) = 8.20 * 10^3 N/C
First, let's find the horizontal acceleration experienced by the proton due to the electric field:
The force experienced by the proton due to the electric field can be calculated using the formula F = q * E, where F is the force, q is the charge, and E is the electric field strength. In this case, the charge of a proton is positive and equal to the elementary charge, e = 1.6 * 10^-19 C.
F = q * E = (1.6 * 10^-19 C) * (8.20 * 10^3 N/C) = 1.312 * 10^-15 N
Next, we can use Newton's second law of motion, F = m * a, to find the horizontal acceleration (a) experienced by the proton:
1.312 * 10^-15 N = (mass of the proton) * a
The mass of a proton is approximately 1.67 * 10^-27 kg. Substituting this value, we can solve for the acceleration (a):
a = (1.312 * 10^-15 N) / (1.67 * 10^-27 kg) ≈ 7.84 * 10^11 m/s^2
Now, we can use the equations of motion to find the time interval (t) required for the proton to travel the given horizontal distance (d). The equation we'll use is:
d = v0 * t + (1/2) * a * t^2
Since the initial velocity (v0) of the proton is given as 3.60 * 10^5 m/s, and we assume it starts from rest in the vertical direction, v0 = 0. Therefore, the equation simplifies to:
d = (1/2) * a * t^2
Plugging in the values, we get:
0.04 m = (1/2) * (7.84 * 10^11 m/s^2) * t^2
Simplifying further:
0.08 = 3.92 * 10^11 t^2
Dividing both sides by 3.92 * 10^11:
t^2 = 2.04 * 10^-13
Taking the square root of both sides:
t = 4.51 * 10^-7 s
Therefore, the time interval required for the proton to travel 4.00 cm horizontally is approximately 4.51 * 10^-7 seconds.