An electron moving in a gravitational free space enters a unifrom electric field E with an initial velocity shown in fig.
.
(a) Find the deflection distance h in the field .
(b) Find an expression for the velocity of electron when it just emerage from the field.
(c ) Find an expression for the total deflection distance H at a vertical screen placed at a distance l from the region of uniform field .(Assume that the field abruptly ends outside the field. ).

Explanation

The electron will move with constant velocity in the x-direction. Time taken by the electron to come out of the plate is `t=l/(v_0)`
Hence, `h=(1)/(2)at^(2)=(1)/(2)((eE)/(m))((l)/(v_0))^(2)=((eE)/(2mv_(0)^(2)))l^(2)`
Velocity of electron when it comes out of the plated is
`vec(v)=vec(u)+vec(at)=v_(0)hat i+((eE)/(m))hat j*t`
`=v_(0)hat i+((eE)/(m)hat j)((l)/(v_(0)))=v_(0)hat i+((eEl)/(mv_(0)))hat j`
After coming out of field, the velocity `vecv_(0)`. Further time taken by electrons to hit the screen is `t'=l//v_(0)`. Hence, further displacement of electron outside the electric field is
`y'=v_(y)*t'=((eEl)/(mv_(0)))(l)/(v_(0))=((eEl^(2))/(mv_(0)^(2)))`
Hence `H=h+y'=((3eEl^(2))/(2 mv_(0)^(2)))`.