In an inkjet printer, letters and images are created by squirting drops of ink h

Question

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether ink is squirted onto the paper or not. The ink drops have a mass m 1.0010-11 kg each and leave the nozzle and travel horizontally toward the paper at velocity-25.0 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 Do 1.90 cm, where there is a uniformm vertical electric field with magnitude E 8.20x104 N/C. (Figure 1) Part A If a drops to be deflected a distance = 0.320 mm by the time it reaches the end ofthe deflect on plate what magnitude o charge q must be gren to the dop? Assume that the density of the ink drop is 1000 /m, and ignore the effects of gravity. Express your answer numerically in coulombs.

Explanation / Answer

E = 8.2 * 10^4 N/C
m = 1* 10^-11 Kg

Initial Velocity in Horizontal Direction = 25.0 m/s
Initial Velocity in Vertical Direction = 0 m/s

Horizontal Distance = d0 = 1.9 * 10^-2 m
Horizontal Acceleration = 0

Time taken to cover Horizontal Distance = t

s = u*t
d0 = 25 * t
t = 1.9 * 10^-2 / 25 s
t = 0.00076 s

Vertical Distance/ Deflection d covered in time t = 0.320 * 10^-3 m

s = u*t + 0.5 at^2
0.320 * 10^-3 = 0 + 0.5 * a * 0.00076^2
a =(0.320 * 10^-3) /(0.5 * 0.00076^2)
a = 1108 m/s^2

We know,
F = q * E
m a = q * 8.2 * 10^4
q = (1* 10^-11) * 1108 /(8.2 * 10^4)
q = 1.35 * 10^-13 c

Magnitude of charge given to the drop q = 1.35 * 10^-13 c

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