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

Answer 1

To determine the time interval over which the proton comes to rest, we can use the kinematic equation:

vf^2 = vi^2 + 2aΔx

Where:
- vf is the final velocity of the proton (which is 0 m/s when it comes to rest)
- vi is the initial velocity of the proton in the positive x-direction (unknown)
- a is the acceleration of the proton (due to the electric field, a = qE/m, where q is the charge of the proton and m is the mass of the proton)
- Δx is the distance traveled by the proton (given as 7.00 cm or 0.07 m)

First, we need to find the initial velocity vi. We know that At1 = 0, so we can write:

At1 = vi/a = vi / (qE/m)

Rearranging the equation, we have:

vi = At1 * (qE/m)

Substituting the given values, we have:

vi = (0.07 m) * (1.6 x 10^-19 C) * (6 x 10^5 N/C) / (1.67 x 10^-27 kg)

Calculating this, we find that vi ≈ 1.6 x 10^6 m/s

Now, we can use the kinematic equation to determine the time interval:

0^2 = (1.6 x 10^6)^2 + 2*(qE/m) * (0.07)

Simplifying and rearranging, we have:

0.112 = (3.85 x 10^12 C/kg) * t^2

Solving for t, we get:

t^2 ≈ 0.112 / (3.85 x 10^12 C/kg)

t ≈ √(0.112 / (3.85 x 10^12 C/kg))

Calculating this, we find that t ≈ 1.33 x 10^-8 s

Therefore, the time interval over which the proton comes to rest is approximately 1.33 x 10^-8 seconds.