Let the point at which the charged particle enters the electric field, be origin O (0, 0), then after travelling a horizontal displacement L, it gets deflected by displacement y in vertical direction as it comes out of electric field.
So, co-ordinates of its initial position are x1 = 0 and y1 = 0 and final position on coming out of electric field are
x2 = L and y2 = y
Components of its acceleration are ax = 0 and qy =

F

m

=

qE

m

and of initial velocity are ux = vx and uy = 0
so, by 2nd equation of motion in horizontal direction,
y2−y1 = uxt+

1

2

axt2 or L - 0 = uxt + 0
t =

L

ux

and by 2nd equation of motion in vertical direction,
y2−y1 = uyt +

1

2

ayt2
or y - 0 = 0 +

1

2

.

qE

m.( L vx )2

... (i)
or y =

qEL2

2mvx2

This gives the vertical deflection of the particle at the far edge of the plate.