In an inkjet printer, letters and images are created by squirting drops of ink h
ID: 2045105 • Letter: I
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 = 1.00×10^-11 each and leave the nozzle and travel horizontally toward the paper at velocity = 21.0m/s . The drops pass through a charging unit that gives each drop a positive charge by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length = 1.90 cm, where there is a uniform vertical electric field with magnitude = 8.10×10^4 N/C
Part A
If a drop is to be deflected a distance = 0.350 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 1000 , and ignore the effects of gravity.
Express your answer numerically in coulombs.
please include number calculations, because i tried but it kept giving wrong answer
Explanation / Answer
F = q*E By Newton's second law, F = m*a, so if the particle has mass m, it will undergo a constant acceleration of: a = q*E/m The equation of motion for a particle undergoing constant acceleration is: delta-y = 0.5*a*t^2 = (q*E/(2*m))*t^2 (2*m*delta-y)/(E*t^2) = q where delta-y is the change in position at time t, from the starting position. There are no forces acting on the droplet in the direction of its initial velocity (parallel to the plates), so the droplet's velocity parallel to the plates is constant. That means it will take: t = 2.4cm/(25 m/s) = 9.6*10^-4 sec for the droplet to traverse the plates. This is the amount of time that the droplet is subjected to the transverse acceleration, so: 2*(1.00*10^-11 kg)*(3.2*10^-4 m)/((8.5*10^4 N/C)*(9.6*10^-4 sec)^2) = q 8.17*10^-14 C = q
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