Q34 of 37 Page 48

Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx= 2.0 × 106 m s–1. If E between the plates separated by 0.5 cm is 9.1 × 102 N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10–19 C, me= 9.1 × 10–31 kg.)

Given,

Velocity of the electron, vx = 2.0 × 106 m s–1


Separation of plates, l = 0.5cm = 0.5 × 10-2 m


Electric field between the plates, E = 9.1 × 102 N C-1


Charge on the electron, q = |e| = 1.6 × 10-19 C


Mass of the electron, me = 9.1 × 10-31 kg


The vertical deflection of the particle (s) is given by





L = √(2.5 × 10-4) m


hence, L = 1.58 × 10-2m = 1.58cm


Hence, the electron will strike the upper plate after travelling 1.58cm.


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