The uniform electric field intensity between the two plates of a parallel plate capacitor is `1 xx 10^(3) Vm^(-1)` acting vertically upwards as shown in the figure.
The plates are sufficiently long and have separation 2 cm. A particle of negative charge `1 mu C` and mass 2 g is projected at an angle `45^(@)` with the electric field from the lower plate with a velocity 'u' . The maximum velocity acquired by the particle , if it is not hit the upper plate is

Options

A. `2 ms^(-1)`
B. `1 ms^(-1)`
C. `0 . 1 ms^(-1)`
D. `0 .2 ms^(-1)`

Answer

D

Explanation

Acceleration of projected charge in y -direction
`a_(y) = -(F)/(m) = - (qE)/(m)`
`= (1 xx 10^(-6) xx 1 xx 10^(3))/(2 xx 10^(-3)) = -(1)/(3) ms^(-2)`
Let projection velocity is u then its vertical component
`u_(x) = ucos theta = u xx cos 45^(@) = (u)/(sqrt(2)) ms^(-1)`
If particle is not to hit upper plate then
`u_(y)^(2) le 2a_(y) (y)`
`rArr (u^(2))/(2) le 2 xx (1)/(2) xx 2 xx 10^(-2)`
`rArr u^(2) le 4 xx 10 ^(-2)`
or `u le 2 xx 10^(-1)`
or ` u le 0 . 2 ms^(-1)`