A particle having charge that of an electron and mass `1.6xx10^(-30)`kg is projected with an initial speed u at the angle `45^(@)` to the horizontal from the lower plate of a parallel-plate capacitor as shown in fig. The plates are sufficiently long and have a separation of 2 cm, find the maximum value of the velocity of he particle so that it does not hit the upper plate. Take the electric field between the plates as `10^(3)Vm^(-1)` directed upward.

Options

A. `2 xx 10^(6)m//s`
B. `2sqrt(2) xx 10^(6)m//s`
C. `sqrt(2) xx 10^(6)m//s`
D. `(1)/(sqrt(2)) xx 10^(6)m//s`

Answer

B

Explanation

`V_(y)^(2) = u_(y)^(2)-2a_(y)s_(y)`
At the highest point `V_(y)` should be 0.
`u_(y)^(2)-2a_(y)s_(y)`
`(u sin 45^(@))^(2)= (2 xx qE)/(m) xx h`
`(u^(2))/(2) = (2qEh)/(m)`
`u = sqrt((4qEh)/(m)) = sqrt((4 xx 1.6 xx 10^(-19)xx 10^(3)xx2 xx 10^(-2))/(1.6 xx 10^(-30)))`
`= 2sqrt(2) xx 10^(6)m//s`