An accelerating voltage of 2.46 * 10^3 V is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the center of a viewing screen 34.0 cm away.
What is the magnitude of the deflection on the screen caused by the Earth's gravitational field?
I am not getting what we have to find here...
thank you
To find the magnitude of the deflection on the screen caused by the Earth's gravitational field, we need to calculate the vertical displacement of the electrons due to the gravitational force.
Here's how you can approach this problem:
Step 1: Determine the horizontal velocity of the electrons:
Since the electrons are traveling horizontally north, there is no horizontal acceleration. Therefore, the horizontal velocity of the electrons remains constant throughout the motion.
Given that the distance to the screen is 34.0 cm and assuming the electrons take t seconds to reach the screen, we can use the formula: distance = velocity * time.
The distance traveled horizontally is 34.0 cm, and the horizontal velocity is v_h. Therefore, we have the equation: 34.0 cm = v_h * t.
Step 2: Calculate the vertical displacement due to the gravitational force:
The vertical displacement (d_y) can be obtained by using the equation: d_y = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time taken to reach the screen.
Step 3: Convert the vertical displacement to centimeters:
Since the other values are given in centimeters, it would be convenient to express the vertical displacement in centimeters as well. Since 1 meter equals 100 centimeters, multiply the displacement (d_y) by 100 to convert it to centimeters.
Step 4: Find the magnitude of the deflection:
The magnitude of the deflection is the absolute value of the vertical displacement (|d_y|).
By following these steps, you should be able to calculate the magnitude of the deflection on the screen caused by the Earth's gravitational field.