an electron enters the region of a uniform,electric field with 3X10^6m/s and e=200 N/C . the horizontal length of the plates is 0.100m. a) find the acceleration of the electron while it is in the electric field . b) if the electron enters the field at time t=0 find the time at which it leaves the field c) the vertical position of the electron as it enters the field is 0 , what is its vertical position when it leaves the field?
a) Well, we have an electron that enters an electric field... Sounds electrifying! Anyway, to find the acceleration of the electron, we can use the formula a = E / m, where E is the electric field strength and m is the mass of the electron. However, you didn't provide the mass of the electron. It's time to call the "Missing Mass Department" to find that out!
b) If the electron entered the field at t = 0, then it must have heard about the new hot spot in town and decided to make a grand exit. To figure out the time it takes for the electron to leave the field, we need to know its initial velocity and the acceleration it experiences in the field. Since you didn't give us its initial velocity, let's just assume it left in a hurry, like it had a date with another electron or something.
c) You mentioned that the vertical position of the electron as it enters the field is 0. Well, I guess it really wants to go "down-to-earth" and have a change in scenery! When it leaves the field, its vertical position depends on the trajectory it follows. Without any additional information, we can't determine its exact vertical position. But hey, let's hope it had nice views during its electric field adventure!
a) To find the acceleration of the electron while it is in the electric field, we can use the formula:
acceleration = electric field / charge of the electron
Given:
Electric field (E) = 200 N/C
Charge of the electron (e) = 1.6 x 10^-19 C
Substituting the values, we have:
acceleration = 200 N/C / 1.6 x 10^-19 C
acceleration = 1.25 x 10^20 m/s^2
Therefore, the acceleration of the electron while it is in the electric field is 1.25 x 10^20 m/s^2.
b) To find the time at which the electron leaves the field, we can use the formula:
time = distance / velocity
Given:
Horizontal length of the plates (d) = 0.100 m
Velocity of the electron (v) = 3 x 10^6 m/s
Substituting the values, we have:
time = 0.100 m / (3 x 10^6 m/s)
time = 3.33 x 10^-8 s
Therefore, the time at which the electron leaves the field is 3.33 x 10^-8 seconds.
c) The vertical position of the electron when it leaves the field can be determined using the formula for vertical motion with constant acceleration:
final_position = initial_position + initial_velocity * time + (1/2) * acceleration * time^2
Given:
Initial vertical position (y0) = 0 (as mentioned in the question)
Initial vertical velocity (vy0) = 0 (since the electron only moves horizontally in the uniform electric field)
Acceleration (a) = acceleration of the electron in the electric field = 1.25 x 10^20 m/s^2 (as calculated in part a)
Time (t) = 3.33 x 10^-8 s (as calculated in part b)
Substituting the values, we have:
final_position = 0 + 0 * 3.33 x 10^-8 + (1/2) * 1.25 x 10^20 * (3.33 x 10^-8)^2
final_position = 0 + 0 + (1/2) * 1.25 x 10^20 * 1.11 x 10^-15
final_position ≈ 0.693 m
Therefore, the vertical position of the electron when it leaves the field is approximately 0.693 meters.
a) To find the acceleration of the electron in the electric field, we can use the formula for the force experienced by a charged particle in an electric field:
F = qE
where F is the force on the electron, q is its charge, and E is the electric field strength.
In this case, the charge of an electron is given as e = 1.6 x 10^-19 C, and the electric field strength is E = 200 N/C.
So, the force experienced by the electron is:
F = (1.6 x 10^-19 C) * (200 N/C)
F = 3.2 x 10^-17 N
According to Newton's second law, the force experienced by an object is equal to its mass multiplied by its acceleration:
F = ma
Rearranging the equation, we can solve for the acceleration:
a = F/m
The mass of an electron is approximately 9.1 x 10^-31 kg.
a = (3.2 x 10^-17 N) / (9.1 x 10^-31 kg)
a = 3.52 x 10^13 m/s^2 (rounded to two significant figures)
Therefore, the acceleration of the electron in the electric field is approximately 3.52 x 10^13 m/s^2.
b) To find the time at which the electron leaves the field, we need to determine the distance it travels while in the field. We can use the formula for the distance traveled during uniform acceleration:
d = (1/2)at^2
where d is the distance traveled, a is the acceleration, and t is the time.
In this case, the initial velocity of the electron is given as 3 x 10^6 m/s, and the acceleration is 3.52 x 10^13 m/s^2.
Since the initial velocity is along the horizontal direction and the plates are horizontal, the vertical position is unaffected. Thus, the vertical position when it leaves the field will be the same as when it entered, which is 0.
c) The vertical position of the electron when it leaves the field is 0.
charge of electron = 1.6 * 10^-19 Coulombs
so force on electron = 200 Newtons / Coulomb * 1.6 * 10^-19 Coulombs
= 3.2 * 10^-17 Newtons
Now it is a freshman Physics problem
F = m a
3.2 * 10^-17 Newtons = 9.11*10^-31 kg * a
a= (3.2/9.11) 10^14 m/s^2 = 3.51*10^13 m/s^2 (part a)
well if it goes 0.1 meters at 3*10^6 m/s it takes (0.1 / 3) *10^-6 seconds
= 3.33 * 10^-8 seconds (hard to time with stopwatch)
h = (1/2) a t^2 = 1.76*10^13 * 11.1 *10^-16 = 19.5 *10^-3 or about 20 millimeters or 2 centimeters (less than an inch)