Electrons in a narrow beam travelling at a steady speed of 5 x 10^7 ms^-1 are directed into the space between two oppositely charged parallel plates 50mm apart. The potential difference between the plates is 2500V. The initial direction of the beam is parallel to the plates. Calculate:

Question

Electrons in a narrow beam travelling at a steady speed of 5 x 10^7 ms^-1 are directed into the space between two oppositely charged parallel plates 50mm apart. The potential difference between the plates is 2500V. The initial direction of the beam is parallel to the plates. Calculate:

The speed and direction of the electrons on leaving the plates.

(The force on each electron between the plates is 8 x 10^-15N and the time each electron spends between the plates if the plates are each 100mm in length is 2 Nanoseconds).

Answer 1

ummm. No.

The force on the electrons is Eq= 2500*1.6e-19 = 4e-16.
x direction: v is constant, i.e. 5e7
y direction: v=at = Ft/m = 4e-16*2e-9/m (sorry don't know the mass of an electron offhand).
Then Pythagorean to find total v and tan-1 to find angle.

Answer 2

To calculate the speed and direction of the electrons on leaving the plates, we need to analyze the force acting on the electrons between the plates.

First, let's calculate the acceleration experienced by each electron. The force on an electron is given by Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the force is 8 x 10^-15 N, and the mass of an electron is approximately 9.11 x 10^-31 kg.

F = m * a

Rearranging the equation to solve for acceleration:

a = F / m

Substituting the given values:

a = (8 x 10^-15 N) / (9.11 x 10^-31 kg)

Calculating the acceleration:

a ≈ 8.77 x 10^15 m/s^2

Now let's find the velocity (speed and direction) of the electrons as they exit the plates.

We know the distance between the plates is 50 mm, which is equivalent to 0.05 m. The time each electron spends between the plates is given as 2 nanoseconds, which is equivalent to 2 x 10^-9 seconds.

Using the kinematic equation:

v = u + a * t

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Since the electrons are initially traveling at a steady speed of 5 x 10^7 m/s, the initial velocity (u) is 5 x 10^7 m/s.

Substituting the known values into the equation:

v = (5 x 10^7 m/s) + (8.77 x 10^15 m/s^2) * (2 x 10^-9 s)

Calculating the final velocity:

v ≈ 5 x 10^7 m/s + 17.54 m/s

v ≈ 5 x 10^7 m/s

Hence, the speed of the electrons on leaving the plates is approximately 5 x 10^7 m/s. Since the initial direction is parallel to the plates, the direction of the electrons remains parallel as they exit the plates.