An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.6 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 47 mm horizontally.

Question

An electron with a speed of 1.1 × 107 m/s moves horizontally into a region where a constant vertical force of 3.6 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 47 mm horizontally.

Answer 1

I will be happy to critique your work. Find the time it takes to move 47mm, then, vertically, d=1/2 a t^2 where a=Force/mass

Answer 2

To determine the vertical distance the electron is deflected, we can use Newton's second law of motion and the equation of motion for uniform acceleration.

First, let's find the acceleration of the electron in the vertical direction using Newton's second law:

Force = mass * acceleration

The force acting on the electron is given as 3.6 × 10^-16 N, and the mass of the electron is 9.11 × 10^-31 kg. Therefore,

3.6 × 10^-16 N = (9.11 × 10^-31 kg) * acceleration

Solving for acceleration, we get:

acceleration = (3.6 × 10^-16 N) / (9.11 × 10^-31 kg)
acceleration ≈ 3.95 × 10^14 m/s²

Now, let's find the time it takes for the electron to cover the horizontal distance of 47 mm (which is equivalent to 0.047 m) using the equation of motion:

distance = initial velocity * time + (1/2) * acceleration * time²

Substituting the given values:

0.047 m = (1.1 × 10^7 m/s) * time + (1/2) * (3.95 × 10^14 m/s²) * time²

This is a quadratic equation in terms of time. Rearranging the terms and setting the equation equal to zero, we get:

(1/2) * (3.95 × 10^14 m/s²) * time² + (1.1 × 10^7 m/s) * time - 0.047 m = 0

Now, we can solve this quadratic equation to find the time.

Using the quadratic formula:

time = (-b ± √(b² - 4ac)) / (2a)

where a = (1/2) * (3.95 × 10^14 m/s²), b = (1.1 × 10^7 m/s), and c = -0.047 m.

Plugging in these values and solving, we find two values for time:

time ≈ 2.82 × 10^-17 s and time ≈ -1.68 × 10^-5 s

Since time cannot be negative, we discard the negative value.

Now, we can calculate the vertical distance using the time.

vertical distance = (1/2) * acceleration * time²

Substituting the values:

vertical distance = (1/2) * (3.95 × 10^14 m/s²) * (2.82 × 10^-17 s)²

vertical distance ≈ 2.81 × 10^-32 m

Therefore, the vertical distance the electron is deflected during the time it has moved 47 mm horizontally is approximately 2.81 × 10^-32 meters.