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A beam of electrons, moving with avelocity of 10^7m/s, enters midway between two horizontal parallel plates in a directions parallel to the plates. Each plate is 5 cm long. These plates are kept 2 cm apart and a potential difference of 90 V is applied between, them. Calculate the velocity of the electron- beam with which it just grazes the edge of the positive plate. (e  /m=1.8×10^{11}  c/kg)
Asked by arjunsah797 | 18 Mar, 2022, 12:06: PM
when electron beam enters the region between parallel plates , it is subjected to electric field due to electrical
potential difference applied between parallel plates .

Hence it is accelerated in vertical direction and gains velocity in vertical direction.

Magnitude of Force acting on electron , F = e E = e ( V/d)

where e is the charge on electron,  E is electric field intensity, V is potential difference and d is distance between plates

acceleration = force / mass = (e/m) ( V/d ) = 1.8 × 1011 × ( 90 / 0.02 ) = 8.1 × 1014 m/s2

Time t for electron to travel the distance that equals the length of plate = distance / velocity

t = 0.05 / 107 = 5 × 10-9 s

velocity vy gained in vertical direction when electron beam grazes the edge of positive plate is
determined from the following equation

vy = a × t = 8.1 × 1014 × 5 × 10-9 = 0.405 × 107 m/s

magnitude of velocity of electron when electron beam grazes the edge of positive plate is given as

Answered by Thiyagarajan K | 18 Mar, 2022, 02:08: PM

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