Inkjet printers can be described as either continuous or drop-on-demand. In a co
ID: 1575114 • Letter: I
Question
Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.6×108 g . The drops will leave the nozzle and travel toward the paper at 50 m/s,passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2 cm long, where there is a uniform vertical electric field with magnitude 9.0×104 N/C . Your team is working on the design of the charging unit that places the charge on the drops.
Part A If a drop is to be deflected 0.25 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? Express your answer with the appropriate units Value Units Submit Previous Answers Request Answer Incorrect; Try Again; 5 attempts remaining Part B How many electrons must be removed from the drop to give it this charge? electrons Submit est Answer Part C f the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 m/s, what q value is needed to achieve the same 0.25-mm deflection? Express your answer with the appropriate units. Value Units Submit Request AnswerExplanation / Answer
a)Time required reaching the end of the plates,
t= X/vix = 0.02/50 = 0.0004s
a= F/m = qE/m - mg = q(9.0*104)/(1.6*1011) – (1.6*1011*9.8) = q(5.625*10^15) -1.568*10^-10
Vertical deflection,
y = viy*t +1/2at^2
y = viy*t +1/2*[(qE)/m]*(0.0004)^2
Put values,
0.00025 = 0*0.0004 +1/2*[ q(5.625*10^15) -(1.568*10^-10)]*(0.0004)^2
q(5.625*10^15) = 3125
q=3125/(5.625*10^15)
q= 5.56*10^-13 C
b) n = q/e = (5.56*10^-13)/(1.6*10^-19) = 3.472*10^6 electrons
c) t= 0.02/25 = 0.0008s
0.00025 = 0*0.0008 +1/2*[ q(5.625*10^15) -(1.568*10^-10)]*(0.0008)^2
q(5.625*10^15) = 625
q= 625/(5.625*10^15)
q= 1.11*10^-13 C
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