In the figure a uniform, upward electric field of magnitude 1.90 × 103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 19.0 cm and separation d = 1.60 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity of the electron makes an angle θ = 48.0° with the lower plate and has a magnitude of 6.40 × 106 m/s. (a) Will the electron strike one of the plates? (b) If so, which plate, top or bottom? (c) How far horizontally from the left edge will the electron strike?
you know the horizontal and vertical components of velocity.
so you have to figure the time to reach the lower plate.
d=vivertical*time+1/2 Force*t^2/mass
force = Ee
d=initial distance to bottom plate.
figure time t. Now, figure the time to get through the maze (.19m/vhorizontal)
did the electron get through or not? If not, what was its horizontal postion at impact time to the lower plate?
To determine whether the electron will strike one of the plates, we need to analyze its motion within the electric field.
(a) To find out if the electron will strike one of the plates, we need to determine its vertical displacement when it reaches the plane of the plates. We can use the equations of motion with constant acceleration to calculate the vertical displacement.
First, let's find the initial vertical velocity component (Vy_0) of the electron:
Vy_0 = V_initial * sin(theta)
Vy_0 = (6.40 × 10^6 m/s) * sin(48.0°)
Vy_0 ≈ 4.83 × 10^6 m/s
Using the equation of motion: Δy = Vy_0 * t + (1/2) * a * t^2
Where Δy is the vertical displacement, t is the time of flight, and a is the acceleration due to the electric field.
Since we have a uniform electric field, the electron will experience a constant acceleration due to the electric field:
a = q * E / m
where q is the charge of the electron, E is the magnitude of the electric field, and m is the mass of the electron.
The charge of the electron, q, is 1.6 × 10^-19 C and the mass, m, is approximately 9.11 × 10^-31 kg.
Substituting the values, we get:
a = (1.6 × 10^-19 C) * (1.90 × 10^3 N/C) / (9.11 × 10^-31 kg)
a ≈ 3.33 × 10^12 m/s^2
Now, we can find the time of flight, t:
Let's consider that when the electron hits the upper plate, its displacement in the y-direction would be equal to the separation, d, between the plates.
Δy = d
Vy_0 * t + (1/2) * a * t^2 = d
We solve this quadratic equation for t. Rearranging and substituting the values:
(1/2) * a * t^2 + Vy_0 * t - d = 0
(1/2) * (3.33 × 10^12 m/s^2) * t^2 + (4.83 × 10^6 m/s) * t - (1.60 × 10^-2 m) = 0
By solving the quadratic equation, we can determine the time of flight, t.
(b) If the time of flight is positive, it means the electron reaches the upper plate first. If the time of flight is negative, it implies the electron reaches the bottom plate first.
(c) To determine the horizontal distance at which the electron strikes, we can use the equation:
x = V_initial * cos(theta) * t
Substituting the values, we can obtain the horizontal distance, x, at which the electron strikes.
By following these steps, you can find the answers to parts (a), (b), and (c) of the question.