At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.5 × 105 m/s and vy = 3.0 × 103 m/s. Suppose the electric field between the plates is given by In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its x coordinate has changed by 2.0 cm?

Question

At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.5 × 105 m/s and vy = 3.0 × 103 m/s. Suppose the electric field between the plates is given by In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its x coordinate has changed by 2.0 cm?

Answer 1

You have left out an important number after

"Suppose the electric field between the plates is given by ..."

You also need to say whether x or y is perpendicular to the places.

Answer 2

*is given by E=(3000i-600j) N/C

the question doesnt include whether x or y is perpendicular to the places

Answer 3

To find the electron's acceleration in the electric field and its velocity when its x coordinate has changed, we will use the given information about the electron's velocity components and the electric field between the plates.

(a) The acceleration of an object in an electric field is given by the equation:

a = q * E / m

where:
a is the acceleration,
q is the charge of the object,
E is the electric field,
m is the mass of the object.

In this case, we are dealing with an electron, so the charge (q) is the charge of an electron which is -1.6 × 10^-19 C (coulombs), and the mass (m) is the mass of an electron which is 9.11 × 10^-31 kg.

The electric field between the plates is given in unit-vector notation, which is E = -8.0 × 10^3 N/C * î, where î is the x-direction unit vector.

Using these values, we can calculate the acceleration of the electron:

a = (q * E) / m
= (-1.6 × 10^-19 C) * (-8.0 × 10^3 N/C) * î / (9.11 × 10^-31 kg)
≈ -1.76 × 10^11 m/s^2 * î

Therefore, the electron's acceleration in the electric field is approximately -1.76 × 10^11 m/s^2 in the x-direction.

(b) To find the electron's velocity when its x-coordinate has changed by 2.0 cm, we can use the equation:

Δx = v * t

where:
Δx is the change in position (given as 2.0 cm),
v is the velocity of the electron in the x-direction,
t is the time.

We need to find the time it takes for the electron to move this distance. Since the motion in the y-direction is not mentioned, we can assume the electron is only moving in the x-direction.

Using the given velocity component vx = 1.5 × 10^5 m/s, the x-velocity of the electron is equal to the velocity of the electron in this case.

Δx = v * t
2.0 cm = (1.5 × 10^5 m/s) * t

Now, we'll solve for t:

t = (2.0 cm) / (1.5 × 10^5 m/s)
= (2.0 × 10^-2 m) / (1.5 × 10^5 m/s)
≈ 1.33 × 10^-7 s

Now that we have the time, we can find the velocity:

v = Δx / t
= (2.0 cm) / (1.33 × 10^-7 s)
= (2.0 × 10^-2 m) / (1.33 × 10^-7 s)
≈ 1.50 × 10^5 m/s

Therefore, the electron's velocity when its x-coordinate has changed by 2.0 cm is approximately 1.50 × 10^5 m/s.