Protons are projected with an initial speed `v_(i)=6kms^(-1)` from a fileds-free region `vec(E ) =-900hat jNC^(-1)` present above the plane as shown in fig. The initial velocity vector of the protons makes an angle u with the plane. The proton are to hit a target that lies at a horizontal cross the plane and enter the electric field. Find the angle `theta` at which the protons must pass through the plane to strike the target.

Explanation

`v_(i)=6.0xx10^(3)ms^(-1)`
`a_(y)=(eE)/(m)=((1.6xx10^(-19))(900))/((1.6xx10^(-27)))=9.0xx10^(10)ms^(-2)`
`R=(v_(i)^(2)sin 2 theta)/(a_(y))=2xx10^(-3)`
`sin 2 theta =(1)/(2)` or `2 theta =30^(@)` or `theta =15^(@), 90^(@)-theta=75^(@)`.