An electron with kinetic energy of 2.010 -16 J is moving to the right along the

Question

An electron with kinetic energy of 2.010-16 J is moving to the right along the axis of acathode-ray tube as shown below. There is an electric field E= (2.0104 N/C) in the region between the deflection plates.Everywhere else, E = 0. (a) How far is the electron from the axis ofthe tube when it reaches the end of the plates?
1____________ mm

(b) At what angle is the electron moving with respect to theaxis?
2______________°

(c) At what distance from the axis will the electron strike thefluorescent screen? _______________cm _______________cm

Explanation / Answer

(a)Let the distance of the electron from the axis of the tubewhen it reaches the end of the plates be d. The force experienced by the electron is F = E * q where E = 2.0 * 104 N/C and q = 1.6 *10-19 C The kinetic energy of the electron is equal to the work doneby the electron in moving to the right along the axis of thecathode-ray tube,therefore we get W = F * d or d = (W/F) where W = 2.0 * 10-16 J (b)Let the angle at which the electron is moving with respectto the axis be . We know from the relation W = V * q * cos or W = E * d * q * cos or cos = (W/E * d * q) or = cos-1(W/E * d * q) ----------(1) (c)Let the distance from the axis where the electronwill strike the fluorescent screen be S.Therefore,we get S = d * The value of obtained from equation (1) will be indegrees.To convert it into radians we multiply with * (/180) radians = * 0.01744 radians

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