If a drop is to be deflected a distance of 0.270 mm by the time it reaches the e

Question

If a drop is to be deflected a distance of 0.270 mm by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is 1000kg/m3).....Answer should be in unit of C

Problem 21.86: Operation of an Inkjet Printer In an inkjet printer, letters are built up by squirting drops of ink at the 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 radius of 14.0 m and leave the nozzle and travel toward the paper at a velocity of 22.0 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 2.05 cm where there is a uniform vertical electric field with a magnitude of 7.70x104 N/C le position whether ink is sq field with a magnitude of 7.70 104 N/C

Explanation / Answer

here,

radius of drop = 14um = 14*10^-6 m
Velocity of drop = 22 m/s
Do = 2.05cm = .00205 m
E = 7.78*10^4 N/C

d = 0.27 mm = 0.00027 m
density of drop = 1000 Kg/m^3

now
Motion parallel to plates
Do = V / T
T = Do / V -----------(1)

Motion perpendicular to plates
d = at^2 / 2 ---------------(2)

using 1 in 2,we get

d = a(Do/v)^2 / 2 -----------(3)

also,The ink drop will be accelerated by the electric field between the plates:

a = F/m
a = qE/m -----------------------(4)

using 4 in 3 we get
q = 2*m*d*v^2 / (E*Do^2)
q = 2 * 10^-11 * 0.000278 * 22^2 / (7.78*10^4 * .00205^2 )

Q = 8.23 * 10^-12 C

magnitude of charge must be given to the drop is 8.23*10^-12 C

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