Two long parallel plates A and B are separated by a distance of 4 cm with an electric field of `45 . 5 Vm^(-1)` between the plates normally from plate A to plate B , as shown in the figure. An electron is projected from plate A with velocity v at an angle of `30^(@)` with the surface of plate A . The maximum value of v so that the electron does not hit plate B is
(Assume gravity free space, charge of electron ` = 1.6 xx 10^(-19)` C and mass of electron ` = 9 . 1 xx 10^(-31) ` kg )

Options

A. `400 km s^(-1)`
B. `3200 km s^(-1)`
C. `800 km s^(-1)`
D. `1600 km s^(-1)`

Answer

D

Explanation

Given, electric field, ` E = 45 . 5 Vm^(-1)`
charge of electron `q_(e) = 1.6 xx 10 ^(-19) C `
mass of electron , `m_(e) = 9 . 1 xx 10 ^(-31) kg` < br> According to the question,
Maximum height in a projectile motion,
`H = (v^(2) sin^(2) theta)/(2a) rArr (4)/(100) = (v^(2)(sin30)^(2))/(2a) " ". . . (i)` (`:. H = 4 cm =(4)/(100)m`)
force in electric field, `F = q_(e) E`
or `m_(e) a = q_(e) E or a= (q_(e)E)/(m_(e))" ". . . (ii)`
From Eqs. (i) and (ii), we get
`:. (4)/(100) = (u^(2) m_(e))/(2(4)q_(e)E) or v = sqrt(((4xx8)/(100)) (q_(e)E)/(m^(e)))`

`v = sqrt((4 xx8)/(100)xx(1.6 xx10^(-19)xx 45.5)/(9.1 xx 10 ^(-31)))= 1600 kms^(-1)`
Hence, the maximum value of v so that the electron does not hit plate b is 1600 `kms^(-1)`