A. A uniform, upward-pointing electric field E of magnitude 3.00�~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 = 4 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v0 of the electron makes an angle ƒÆ=45�‹ with the lower plate and has a magnitude of 6.93�~106 m/s. Will the electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).

B.The next electron has an initial velocity which has the same angle ƒÆ=45�‹ with the lower plate and has a magnitude of 5.83�~106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates ( m).

Can you express your angles some other way that

ƒÆ=45�‹ ?

I can't make sense of that

sorry its theta=45 degrees for both questions

To determine whether the electron will strike one of the plates or leave the space between the plates, we need to analyze the motion of the electron in the electric field.

Let's break down the problem:

A. Initial conditions:
- Electric field magnitude: E = 3.00 × 10^3 N/C (pointing upward)
- Plate length: L = 4 cm (or 0.04 m)
- Plate separation: d = 2.00 cm (or 0.02 m)
- Initial velocity magnitude: v0 = 6.93 × 10^6 m/s
- Initial angle: φ = 45° (with respect to the lower plate)

We can start by calculating the acceleration experienced by the electron due to the electric field. The electric force on the electron is given by:

F = q * E

where q is the charge of the electron. Since an electron has a negative charge (e = -1.6 × 10^-19 C), the force will be downward. Therefore, we need to consider the magnitude of the electric field:

F = q * |E|

The mass of an electron is m = 9.11 × 10^-31 kg. We can calculate the acceleration (a) using Newton's second law:

a = F / m

Plugging in the values:

a = (q * |E|) / m

Now we can calculate the acceleration magnitude:

a = (e * |E|) / m

Next, we need to determine the time it takes for the electron to travel between the plates. Assuming the motion is purely horizontal (as the electric field is vertical), we can use kinematic equations.

The horizontal distance traveled (x) can be calculated using the equation:

x = v0 * t

where t is the time.

To find t, we need to determine the time it takes to travel the vertical distance (y) between the plates. Using the equation:

y = 0.5 * a * t^2

and since the initial vertical velocity (v0y) is 0 (as the initial angle is with respect to the horizontal):

y = 0.5 * a * t^2

We can solve this equation for t:

t = sqrt((2 * y) / a)

If the time t is greater than time taken for the electron to move horizontally, the electron will strike one of the plates. Otherwise, it will leave the space between the plates.

To find the horizontal distance from the left edge of the lower plate, we can substitute the value of t into the equation:

x = v0 * t

We can now perform this analysis for both parts of the problem:

A. For the given initial conditions:

- Calculate acceleration: Calculate a using a = (e * |E|) / m
- Calculate time to travel vertically: t = sqrt((2 * y) / a)
- Calculate horizontal distance: x = v0 * t
- If x is greater than L (0.04 m), the electron strikes one of the plates. Otherwise, record the vertical position as the answer.

B. Repeat the same steps for the given initial conditions in part B.

By calculating the above values and comparing them to the given conditions, we can determine whether the electrons will strike the plates or leave the space between them.