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CBSE Class 12-science Answered

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
Expert Answer
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
begin mathsize 14px style v space equals space square root of v subscript x superscript 2 plus v subscript y superscript 2 end root space equals 10 to the power of 7 space square root of 1 plus left parenthesis 0.405 right parenthesis squared end root space m divided by s space equals space 1.079 space cross times 10 to the power of 7 space end exponent m divided by s end style
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