Question

The electron beam inside an old television picture tube is 0.40 mm in diameter and carries a current of 50 μA. This electron beam impinges on the inside of the picture tube screen.

Part A. What is the current density in the electron beam?
Express your answer with the appropriate units.
Part B. The electrons move with a velocity of 3.8 ×107 m/s. What electric field strength is needed to accelerate electrons from rest to this velocity in a distance of 5.0 mm?
Express your answer with the appropriate units.
Part C. Each electron transfers its kinetic energy to the picture tube screen upon impact. What is the powerdelivered to the screen by the electron beam?
Express your answer with the appropriate units.

Answer 1

The current density in the electron beam is approximately 3.98 x 10⁵ A/m². The electric field strength required to accelerate electrons to the given velocity is about 1.72 x 10⁴ V/m. The power delivered to the screen by the electron beam is around 4.29 x 10⁻³ W.

Step-by-step explanation:

Part A: Current Density Calculation;

The current density J can be calculated using the formula J = I/A, where I is the current and A is the cross-sectional area of the beam. Given a current I of 50 µA and a beam diameter of 0.40 mm, the radius r is 0.20 mm (or 0.20 x 10-3 m). The area A is πr², which equates to π(0.20 x 10-3 m)2. Using these values:

J = 50 x 10-6 A / (π(0.20 x 10-3 m)²)

After calculating, the current density J is approximately 3.98 x 105 A/m².

Part B: Electric Field Strength Calculation:

To calculate the electric field strength E needed to accelerate electrons from rest to a velocity v over a distance d, we use the equation eE = m(v²)/(2d), where e is the charge of an electron (1.60 x 10-19 C), m is the mass of an electron (9.11 x 10-31 kg), v is the velocity of the electrons (3.8 x 107 m/s), and d is the distance (5.0 mm or 5.0 x 10-3 m). The calculation gives us:

E = (9.11 x 10-31 kg)(3.8 x 107 m/s)² / (2 x 1.60 x 10-19 C x 5.0 x 10-3 m)

The required electric field strength E is approximately 1.72 x 104 V/m.

Part C: Power Delivered to the Screen Calculation:

To find the power delivered to the screen P, use the formula P = IV, where I is the current and V is the potential difference the electrons traverse. The kinetic energy (KE) of an electron with velocity v is given by KE = (1/2)mv², and since one electron volt (eV) equals 1.60 x 10-19 J, we can determine the potential difference V in volts by equating KE to eV:

V = (1/2)mv²/e

P = I (1/2)mv²/e

Substituting known values:

P = 50 x 10-6 A (1/2)(9.11 x 10-31 kg)(3.8 x 107 m/s)²/ (1.60 x 10-19 C)

The power P is approximately 4.29 x 10-3 W.

Answer 2

A) J= 398 A/m²

B) E= 1.6×10⁶ N/C

C) P= ×10⁴ W

Step-by-step explanation:

My work is in the attachment. Comment with questions or if something seems wrong with my work. (Honestly, they seem little high but it could just be the given numbers being unrealistic.) Below I have explanations of each part to match up with the image as well.

Part A:

Current density (J) is defined as the amount of current in a particular cross-sectional area. To get this, we simply need to divide the current (I) by the cross-sectional area of the electron beam tube (A).

Part B:

This one took the most work for me. I used a kinematic equation (yes they apply to electrons) to find the electric field (E). I used a modified form of the familiar: ∆d=V₀τ+aτ²/2

We can use the fact that τ= V/a, a=(qE/m), and V₀=0 here to rewrite the equation in terms of values we know and/or can look up. From there we solve for E and plug in the values.

Part C:

Power (P) is simply work (W) over time (τ). We know what τ is from before and can take W= mV²/2. Plugging these in and reducing some values gives us an equation for power as well.