Question

An electron is launched from the negative plate It strikes the positive plate at a speed of 2.50*107 m(s What was the electron's speed as it left the negative plate? Express your answer to two significant figures and include the appropriate units. View Available Hint(s) Value Units Submit

Answer 1

To find the electron's initial speed, use the equation for acceleration due to an electric field. Calculate the acceleration using the known values, then use a kinematic equation to find the initial speed.

Step-by-step explanation:

To find the electron's initial speed, we need to use the equation for acceleration due to an electric field: a = qE/m, where q is the charge of the electron, E is the electric field strength, and m is the mass of the electron. We know that the electron's charge is -1.6 x 10-19 C and the electric field strength is 2.0 x 105 N/C. The mass of the electron is 9.11 x 10-31 kg. Plugging in these values, we can calculate the acceleration: a = (-1.6 x 10-19 C)(2.0 x 105 N/C) / (9.11 x 10-31 kg). Solving this equation gives us the acceleration of the electron.

Once we have the acceleration, we can use the kinematic equation vf = vi + at to find the initial speed of the electron. We know that the final speed is 2.50 x 107 m/s and the time is the distance between the negative and positive plates divided by the initial speed. Plugging in these values, we can solve for the initial speed.

Answer 2

The electron's speed as it left the negative plate was 2000 m/s.

Step-by-step explanation:

The electron's speed as it left the negative plate can be found using the conservation of energy and the potential difference between the plates.

Given that the electron's speed as it strikes the positive plate is 2.50*107 m/s and the distance between the plates is 2 cm, we can calculate the electric field strength using the formula:

E = ΔV / d

Where E is the electric field strength, ΔV is the potential difference, and d is the distance between the plates.

Substituting the given values:

E = 200 V / 0.02 m = 10000 N/C

Since the electric field accelerates the electron, we can use the equation of motion:

v2 = u2 + 2ad

Where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance traveled.

Since the electron starts from rest at the negative plate, the initial velocity is 0 m/s.

Substituting the values into the equation:

2.50*107 m/s2 = 0 + 2 * 10000 * 0.02

Simplifying the equation:

v = sqrt(2 * 10000 * 0.02) = 2000 m/s

Therefore, the electron's speed as it left the negative plate was 2000 m/s.