Figure shows an assembly of deflecting plates A and B ofan ink-jet printer which causes moving ink droplets to deflect at desired displacements by continuously varying electric field between the plates. An ink drop with a mass `m = 1.3 xx 10^(-10)kg` and a negative charge of magnitude `q = 1.5x x 10^(-13)C` enters the region between the plates, initially moving along the x-axis with speed `v_(x), = 10m//s`. The length of plates is `L = 1.6cm`. The plates are connected with a varying voltage and thus produce an electric field at all points between them. Assume that field `vec(E)` for some duration is constant and it is acting in downward direction as shown and has a magnitude of `E = I .4 xx 10^(6) N//C`, find the vertical deflection of the drop at the far edge ofthe plate? As the gravitational force on the drop is very small relative to the electrostatic force acting on the drop, it can be neglected for this analysis. `[6.4 xx 10^(-4)m]`

Explanation

KEY IDEAS
The drop is negatively charged and the electric field is directed downward. From Eq. 22-45, a constant electrostatic force of magnitude QE acts upward on the charged drop. Thus, as the drop travels parallel to the x axis at constant speed `v_(x)`, it accelerates upward with some constant acceleration `a_(y)`.
Calculations: Applying Newtons.s second law (F = ma) for components along the y axis, we find that
`a_(y)=(F)/(m)=(QE)/(m). " " (22-47)`
Let t represent the time required for the drop to pass through the region between the plates. During t the vertical and horizontal displacements of the drop are
`y=(1)/(2) a_(y)t^(2) and L=v_(x)t," " (22-48)`
respectively. Eliminating t between these two equations and substituting Eq. 22-47 for `a_(y)`, we find
`y=(QEL^(2))/(2mv_(x)^(2))`
`=((1.5xx10^(-13)C)(1.4xx10^(6)N//C)(1.6xx10^(-2)m)^(2))/((2)(1.3xx10^(-10)kg)(18m//s)^(2))`
`=6.4xx10^(-4)m=0.64mm.` (Answer)