38. A proton initially moves left to right long the ´‑axis at a speed of 2 ´ 103 m/s. It moves into an electric field, which points in the negative ´ direction, and travels a distance of 0.2 m before coming to rest. If the proton's mass and charge are 1.67 ´ 10‑27 kg and 1.6 ´ 10‑19 C respectively, what is the magnitude of the electric field?
a. 28.3 N/C
b. 13.9 N/C
c. 0.104 N/C
d. 0.038 N/C
Force=Eq
E= Force/q=massproton*acceleration/q
you can get acceleration from the previous post.
To find the magnitude of the electric field, we can use the following formula:
Electric Force on Proton = Charge of Proton * Electric Field Strength
The electric force on the proton can be calculated using Newton's second law:
Electric Force on Proton = Mass of Proton * Acceleration
Since the proton comes to rest, its final velocity is 0. We can use the kinematic equation:
(vf)^2 = (vi)^2 + 2 * a * d
Where:
vf = final velocity (0 m/s)
vi = initial velocity (2 * 10^3 m/s)
a = acceleration
d = distance (0.2 m)
Substituting the given values, we get:
0 = (2 * 10^3)^2 + 2 * a * (0.2)
Simplifying the equation:
-(2 * 10^3)^2 = 2 * a * (0.2)
Solving for acceleration, we find:
a = -(2 * 10^3)^2 / (2 * 0.2)
Now, let's substitute the values into Newton's second law:
Electric Force on Proton = (1.67 * 10^-27 kg) * a
Finally, substituting this into the electric field formula, we get:
Charge of Proton * Electric Field Strength = (1.67 * 10^-27 kg) * a
Simplifying, we find:
Electric Field Strength = a * (1.6 * 10^-19 C) / (1.67 * 10^-27 kg)
Calculating this expression gives us:
Electric Field Strength ≈ -9.672 * 10^4 N/C
Since the electric field points in the negative direction, we take the magnitude of the value:
|Electric Field Strength| ≈ 9.672 * 10^4 N/C
Therefore, the magnitude of the electric field is approximately 9.672 * 10^4 N/C.
None of the provided answer choices match this value, so please double-check the problem or the answer choices given.