+3 votes
in Physics by kratos

In a television set, electrons are first accelerated from rest through a potential difference in an electron ***. They then pass through deflecting plates before striking the screen.

a. Determine the potential difference through which the electrons must be accelerated in the electron ** in order to have a speed of 6.0 x 107 m/ when they enter the deflecting plates.

The pair of horizontal plates shown below is used to deflect electrons up or down in the television set by placing a potential difference across them. The plates have length 0.04 m and separation 0.012 m, and the right edge of the plates is 0.50 m from the screen. A potential difference of 200 V is applied across the plates, and the electrons are deflected toward the top of the screen. Assume that the electrons enter horizontally midway between the plates with a speed of 6.0 x 107 m/* and that fringing effects at the edges of the plates and gravity are negligible.

b. Which plate in the pair must be at the higher potential for the electrons to be deflected upward? Check the appropriate box below.

c. Considering only an electron'* motion as it moves through the space between the plates, compute the following.

i. The time required for the electron to move through the plates

ii. The vertical displacement of the electron while it is between the plates

d. Show why it is a reasonable assumption to neglect gravity in part c.

e. Still neglecting gravity, describe the path of the electrons from the time they leave the plates until they strike the screen. State a reason for your answer.

1 Answer

+4 votes
by kratos
 
Best answer

a. W = qV = mv2 gives V = mv2/2q = 1.0 ×104V

b. Electrons travel toward higher potential making the upper plate at the higher potential.

c.

i. x = vxt gives t = 6.7 × 10–10s

ii. F = ma = qE and E = V/d gives a = qV/md and y = at2 (v0y = 0) gives y = qVt2/2md = 6.5 × 10–4 m

d. Fg is on the order of 10–30 N (mg) and FE = qE = qV/d is around 10–14 N so FE ≫ Fg

e. Since there is no more electric force, the path is a straight line.

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