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jekas [21]
4 years ago
5

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lerated by a voltage of 3.0 kV; the beam was then steered to different points on the screen by coils of wire that produced a magnetic field of up to 0.65 T .
What is the speed of electrons in the beam?
Physics
1 answer:
oee [108]4 years ago
3 0

Answer:

3.25 × 10^7 m/s

Explanation:

Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)

Ek = 1/2 mv2 = qΔV .................. 1

Given that V is the electron speed in m/s

Charge of electron = 1.60217662 × 10-19 coulombs

Mass of electron = 9.109×10−31 kilograms

ΔV = 3.0kV = 3000V

Make V the subject of the formula in eqaution 1

V = sqr root 2qΔV/m

V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31

V = 3.25 × 10^7 m/s

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(a) The work done in moving the unit charge from point C to A is 7.62 x 10⁻³ J.

(b) The work done in moving the unit charge from point D to B is 7.62 x 10⁻³ J.

<h3>Work done in moving the charge from C to A</h3>

W = Fd

W = Kq²/d

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  • from 0 origin to A, d = 5 m

W(C to A) = W(0 to C) + W(0 to A)

W(C \ to \ A) = - \frac{Kq^2}{7.07} + \frac{Kq^2}{5} \\\\ W(C \ to \ A)  = 0.0586 \ Kq^2\\\\W(C \ to \ A)  = 0.0586 \times 9 \times 10^9 \times (3.8\times 10^{-6})^2\\\\W(C \ to \ A)  = 7.62 \times 10^{-3} \ J

<h3>Work done in moving the charge from D to B</h3>
  • from 0 origin to D, d = √(5² + 5²) = 7.07 m
  • from 0 origin to B, d = 5 m

W(D to B) = W(0 to D) + W(0 to B)

W(D to B) = 7.62 x 10⁻³ J

Learn more about work done here: brainly.com/question/25573309

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