# Ink Drops Deflection

Deflection of ink drops in a printer by electric field. The ink drops have a mass $m = 1.00*10^{11}$ kg each and leave the nozzle and travel horizontally toward the paper at velocity $v = 16.0$ m/s . The drops pass through a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length $D0 = 1.55 cm$ , where there is a uniform vertical electric field with magnitude $E = 7.90*10^4$ N/C . (Figure 1)

Part A

If a drop is to be deflected a distance $d = 0.290$ mm by the time it reaches the end of the deflection plate, what magnitude of charge q must be given to the drop? Assume that the density of the ink drop is $1000(kg/m^3)$ , and ignore the effects of gravity.

Time to travel the deflection plates is

$t = D/V =0.0155/16 =0.96875$ ms

Acceleration of drop is

$a = F/m = q*E/m$

Total deflection between plates is

$d = a*t^2/2$

Therefore

$2*d =q*(E/m)*(D/V)^2$

$q = (2d)*(m/E)* (V/D)^2 =7.823*10^{-14} C =48900*e (C)$