An electron moves to the right with speed v along the axis of a cathode ray tube

Question

An electron moves to the right with speed v along the axis of a cathode ray tube. There is an electric field E~ = E0 ˆj in the region between the deflection plates, which are of length l, and everywhere else E~ = 0. The flat screen is a distance L from the end of the plates. Assume that the electron is moving fast enough that it will not “fall” or hit the deflection plates while crossing the deflection zone (ignore effect of the gravitional force on the electron as it is negligible across the entire distance). Find ?y, the deflection from the center point where the electron hits the screen. You might want to break the problem up into two parts as the figure hints.

The kinetic energy of the electron is equal to the

work done.

(1/2)mv2 = eVa

v2 = 2[eVa/m]

The electric force exerted on the electron is,

F = eE

ma = eE

Thus, the acceleration of the electron is,

a = eE/m

The time interval is,

  t = l/v

Use the below equation to find the deflection of electron.

what is the total vertical displacement ?y (in m)?

Explanation / Answer

time taken to cross the plates, t1 = l/vx

= 0.05/(5*10^6)

= 1*10^-8 s

acceleration of electron in y-direction,

a = q*E/m

= 1.6*10^-19*1000/(9.1*10^-31)

= 1.758*10^14 m/s^2

distance travelled in in y-direction before crossing the plates,

y1 = 0.5*a*t1^2

= 0.5*1.758*10^14*(1*10^-8)^2

= 8.79*10^-3 m

= 0.00879 m

veloicty gained in y-direction befor crossing the plates,

vy = a*t1

= 1.758*10^14*1*10^-8

= 1.758*10^6 m/s

ramaining time taken to hit the screen,

t2 = L/vx

= 0.2/(5*10^6)

= 4*10^-8 s

distance travelled in this time, y2 = vy*t2

= 1.758*10^6*4*10^-8

= 0.07032 m

so, total dispalcement in y-direction, delta_Y = y1+y2

= 0.00879 + 0.07032

= 0.0791 m <<<<<<<<-------------------Answer

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