Question

If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case

Answer 1

The speed of the electron when it reaches the positive plate is approximately 4.0 × 105 m/s.

Step-by-step explanation:

To find the speed of the electron when it reaches the positive plate, we need to use the principles of electric fields and acceleration.

1. The initial speed of the electron is given as 4.0 × 105 m/s.
2. We are given the acceleration due to the electric field, which is 1 m/s².
3. We can use the equation v^2 = u^2 + 2as to calculate the final speed, where u is the initial speed, a is the acceleration, and s is the distance travelled. In this case, the distance travelled is the separation between the negative and positive plates (d = 2 cm).
4. Substituting the known values:

v^2 = (4.0 × 105 m/s)^2 + 2(1 m/s²)(0.02 m)

1. Simplifying the equation, we get:

v^2 = 1.6 × 1011

1. Taking the square root of both sides of the equation:

v = √(1.6 × 1011) ≈ 4.0 × 105 m/s

Therefore, the speed of the electron will be approximately 4.0 × 105 m/s when it reaches the positive plate.

Answer 2

v = -v₀ / 2

Step-by-step explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

v² = v₀² - 2a y

when the initial velocity is vo it reaches just the negative plate so v = 0

a = v₀² / 2y

now they tell us that the initial velocity is half

v’² = v₀’² - 2 a y’

v₀ ’= v₀ / 2

at the point where turn v = 0

0 = v₀² /4 - 2 a y '

v₀² /4 = 2 (v₀² / 2y) y’

y = 4 y'

y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

v² = v₀² -2a y’

v² = 0 - 2 (v₀² / 2y) y / 4

v² = -v₀² / 4

v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.