A particle of mass m and charge -q enters the region between the two charged plates initially moving along x-axis with speed `v_(x)` as shown in figure. The length of plate is L and a uniform electric field E is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is

Options

A. `(qEL^(2))/(2mv_(x)^(2))`
B. `(qEL^(2))/(2mv_(x))`
C. `(2mv_(x)^(2))/(qEL^(2))`
D. `(2mv_(x))/(qE^(2)L)`

Answer

A

Explanation

Here, in the vertical direction,
initial velocity, v = 0
acceleration, `a=F/m=(qE)/m" "(therefore F=qE)" "...(i)`
Time taken to cross the field, `t=(distance)/(velocity)=L/v_(x)" "...(ii)`
(`therefore` velocity along the horizontal direction is constant)
`therefore" "s=vt+1/2at^(2)`
`therefore" "` Deflection,
`y=0+1/2((qE)/m)(L/v_(x))^(2)`
[Using (i) and (ii)]
`therefore" "y=(qEL^(2))/(2mv_(x)^(2))`