Question

An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 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 29 mm horizontally.

Answer 1

Answer: 7.86×10^-4 mm

Explanation: find the attached file for the solution

Answer 2

y = 77.74 10⁻⁵ m

Step-by-step explanation:

For this exercise we can use Newton's second law

F = m a

a = F / m

a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

x= 29 mm = 29 10⁻³ m

On the x axis

v = x / t

t = x / v

t = 29 10⁻³ / 1.7 10⁷

t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

y =
`v_(oy)` t + ½ a t²

y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

y = 77.74 10⁻⁵ m