In an inkjet printer, letters are built up by squirting dropsof ink at the paper

Question

In an inkjet printer, letters are built up by squirting dropsof ink at the paper from a rapidly moving nozzle. The pattern onthe paper is controlled by an electrostatic valve that determinesat each nozzle position whether ink is squirted onto the paper ornot. The ink drops, 15.9 mm in radius,leave the nozzle and travel toward the paper at 18.1 m/s. The dropspass through a charging unit that gives each drop a positive chargeq when the drop loses some electrons. The drops then passbetween parallel deflecting plates 2 cm in length where there is auniform vertical electric field with magnitude 76000 N/C. If a dropis to be deflected 0.3 mm by the time it reaches the end of thedeflection plate, what magnitude of charge must be given to thedrop? (Assume that the density of the ink drop is the same as thatof water, 1000 kg/m^3.)

Explanation / Answer

time to reach end = distance / speed = 0.02 m / 18.1 =   0.001105 seconds . the deflection is under constant acceleration, so... .      d = (1/2) at2  or         a = 2 d / t2 = 2 * 0.0003 /0.0011052 =    491.415m/s2 . The volume of the drop is   (4/3) r3 = (4/3) * (15.9 x 10-6)3 = 1.68376 x 10-14m3 . So the mass of the drop is   density * volume = 1000 * 1.68376 x 10-14 = 1.68376 x10-11 kg . The force on the drop must be    F = ma = 1.68376 x 10-11 * 491.415 = 8.27426x 10-9 Newtons . and finally... .       F = qE   so     q = F / E = 8.27426 x 10-9 / 76000 =    1.0887 x10-13 Coulomb   is the charge . there's a lot of calculations here... if my answer is notcorrect, please let me know. I may have miscalculated something andI will correct it if necessary. . there's a lot of calculations here... if my answer is notcorrect, please let me know. I may have miscalculated something andI will correct it if necessary.

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