Question

an accelerating voltage of 2.50 3 103 v is applied to an electron gun, producing a beam of electrons originally traveling horizontally north in vacuum toward the cen- ter of a viewing screen 35.0 cm away. what are (a) the magnitude and (b) the direction of the deflection on

Answer 1

Main Answer:

(a) The magnitude of the deflection is given by the equation
`d = (1/2) * a * t^2,`where 'd' is the deflection, 'a' is the acceleration due to the applied voltage, and 't' is the time of flight. (b) The direction of the deflection is perpendicular to the initial velocity, forming a right angle with the northward horizontal direction.

Therefore, the correct answer is a) b).

Step-by-step explanation:

The magnitude of the deflection (d) is determined by the equation
`d = (1/2) * a * t^2,` where 'a' is the acceleration due to the applied voltage. In this scenario, the electron gun applies an accelerating voltage of 2.50 * 10^3 V, causing the electrons to accelerate. The time of flight ('t') is not explicitly provided, but it can be calculated based on the given information. The distance to the viewing screen is 35.0 cm, and assuming the electrons are moving horizontally north initially, the deflection is perpendicular to this direction.

The direction of the deflection is crucial in understanding the motion of the electrons. When an electric field is applied vertically, the electrons experience a force perpendicular to their initial velocity. As a result, the electrons deviate from their original path, forming a right angle with the northward horizontal direction.

In summary, the magnitude of the deflection is determined by the acceleration due to the applied voltage and the time of flight, while the direction is perpendicular to the initial velocity. These factors collectively describe the behavior of the electrons in response to the applied voltage and help visualize the deflection on the viewing screen.

Therefore, the correct answer is a) b).