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 simple steps
To determine the time interval over which the proton comes to rest, we can use the principle of work-energy.
First, we need to calculate the work done on the proton. The work done by the electric field is given by the equation:
W = q * ΔV
where W is the work done, q is the charge of the proton, and ΔV is the change in electric potential. In this case, since the proton comes to rest, the change in electric potential is equal to the initial electric potential.
Therefore, ΔV = -E * Δx
where E is the electric field and Δx is the distance travelled by the proton. Plugging in the values:
ΔV = -(-6x10^5 N/C) * 0.07m = 4.2x10^4 J
Since the work done on the proton is equal to the change in its kinetic energy, we can write:
ΔK = W
Solving for the change in kinetic energy:
ΔK = (1/2) * m * (v^2 - 0^2)
Since the proton comes to rest, the final velocity (v) is zero. Therefore:
ΔK = (1/2) * m * (0 - 0^2)
ΔK = 0
Setting the work done on the proton equal to zero:
0 = 4.2x10^4 J
This tells us that zero work was done on the proton, which means it lost all its initial kinetic energy and converted it into electric potential energy.
Now, let's calculate the initial kinetic energy of the proton:
K = (1/2) * m * v^2
Since it is given that the initial velocity and initial kinetic energy are both zero, we can write:
0 = (1/2) * m * (0^2)
0 = 0, which is true.
Therefore, the time interval over which the proton comes to rest is instantaneous, as it loses all its initial kinetic energy and comes to rest immediately.
To determine the time interval over which the proton comes to rest, we can use the equation of motion for uniformly accelerated linear motion:
vf = vi + at
Here, vf is the final velocity (which is 0 since the proton comes to rest), vi is the initial velocity, a is the acceleration, and t is the time interval.
Given:
Initial velocity, vi = 0 m/s (at rest)
Acceleration, a = E/m, where E is the electric field and m is the mass of the proton
We need to find the time interval, t.
Step 1: Calculate the acceleration using the given electric field
a = E/m
= (-6 * 10^5 N/C) / (1.6 * 10^-19 kg)
≈ -3.75 * 10^24 m/s^2 (rounded to two significant figures)
Step 2: Substitute the values into the equation of motion
0 = 0 + (-3.75 * 10^24 m/s^2) * t
Step 3: Solve for t by rearranging the equation
0 = -3.75 * 10^24 m/s^2 * t
Since any number multiplied by 0 is equal to 0, the equation is always satisfied. This suggests that the proton will never come to rest and will continue to accelerate indefinitely.
Therefore, the time interval cannot be determined as the proton does not come to rest.
To determine the time interval over which the proton comes to rest, we need to use the equations of motion for an object moving in a uniform electric field.
Step 1: Identify the given information:
- Initial velocity of the proton (v₀ = 0)
- Displacement of the proton (s = 7.00 cm = 0.07 m)
- Electric field strength (E = -6 × 10^5 N/C)
Step 2: Determine the acceleration:
The force experienced by the proton in the electric field can be calculated using the equation F = qE, where F is the force, q is the charge of the proton, and E is the electric field strength. For a proton, q = +1.6 × 10^-19 C. Therefore, the force experienced by the proton is:
F = (1.6 × 10^-19 C) × (-6 × 10^5 N/C)
F = -9.6 × 10^-14 N
The force experienced by an object is related to its acceleration by the equation F = ma, where m is the mass of the object and a is its acceleration. The mass of a proton is approximately 1.67 × 10^-27 kg, so we can calculate the acceleration:
-9.6 × 10^-14 N = (1.67 × 10^-27 kg) × a
a ≈ -5.75 × 10^12 m/s^2 (negative sign indicates acceleration opposite to the direction of motion)
Step 3: Use the equations of motion:
The equation to calculate displacement (s) in terms of initial velocity (v₀), final velocity (v), and acceleration (a) is:
s = (v₀ + v) / 2 × t
Since the proton comes to rest, its final velocity (v) is 0.
Substituting the given values:
0.07 m = (0 + 0) / 2 × t
0.07 m = 0 × t
t = 0
Step 4: Interpretation:
From the calculation, we find that the time interval over which the proton comes to rest is 0 seconds. This result indicates that the proton comes to rest instantly in the uniform electric field.