Operation of an Inkjet Printer In an inkjet printer, letters and images are crea

Question

Operation of an Inkjet Printer 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 = 1.00 Times 10_11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 20.0 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 D_o = 1.75 cm. where there is a uniform vertical electric field with magnitude E = 7.55 Times 10^4 N/C If a drop is to be deflected a distance d = 0.250 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. Express your answer numerically in coulombs.

Explanation / Answer

here,
mass of drop, m = 10^-11 kg
velocity of drop, v = 20 m/s
PLate length, do = 1.75 cm = 0.0175 m
Electric field, E = 7.55*10^4 N/C
deflection, d = 0.250mm = 0.250*10^-3 m

Parallel distance tarvelled, Do = v*t ( t is time)
perperndicular distance travelled, d = 0.5*a*t^2

therefore, d = 0.5*a*(Do/v)^2 -----------(1)

The ink drop will be accelerated by electric field betweent he plates, so from newton second law,
F = m*a
a = F/m
a = q*E/m ( since electric force, f = q*E)

From equation 1, and rewriting for charge, q

q = (2*m*d*v^2) / (E(D0)^2)
q = (2*10^-11*0.250*10^-3*20^2) / (7.55*10^4(0.0175)^2)

q = 8.65*10^-14 C

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