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 The ink drops have a mass ( exttip{m}{m}) = 1.00×1011 ({ m kg}) each and leave the nozzle and travel horizontally toward the paper at velocity ( exttip{v}{v}) = 19.0 ({ m m/s}) . The drops pass through a charging unit that gives each drop a positive charge ( exttip{q}{q}) by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length ( exttip{D_{ m 0}}{D_0}) = 1.60 ({ m cm}) , where there is a uniform vertical electric field with magnitude ( exttip{E}{E}) = 7.60×104 ({ m N/C}) If a drop is to be deflected a distance ( exttip{d}{d}) = 0.320 ({ m mm}) by the time it reaches the end of the deflection plate, what magnitude of charge ( exttip{q}{q}) must be given to the drop? Assume that the density of the ink drop is 1000 ({ m kg/m^3}) , and ignore the effects of gravity.

Explanation / Answer

mass of inkdrop, m = 10^-11 kg

velocity, v = 19 m/s

length of plates, l = 1.6 cm = 0.016 m

electric field, E = 7.60*10^4 N/C

distance to drop, d = 0.320 mm = 0.000320 m

From columb's law, electric force is given as,
F = qE

Since F = ma, we also know that:
ma = qE
a = qE/m-------------------------(1)

From second eqn of motion, horizontal distance
x = 1/2at^2
x = 1/2 * (qE/m) * t^2 ---------------------(2)

also, upon firthur solving ,
x = 1/2 * (qE/m) * t^2
2x / (t^2) = qE/m
q = (2 * x * m) / (E * t^2) -------------------(1)

Since the plates are 0.016 m long, the time will be:
t = 0.016 / 19
t = 0.0008421 s

Therefore:
q = (2 * d * m) / (E * t^2)
q = (2 * 0.00032 * (1*10^-11)) / ((7.60*10^4) * (0.0008421^2))
q = 1.188*10^-13 C

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