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 paper or not.

Chrome-MasteringPhysics: Homework 2 Secure | https:// session. masteringphysics.com/myct/itemView?assignmentProblemID-82506681 Homework2 Item 4 Resources « previous | 4 of 6 next» Item 4 The ink drops have a mass m = 1.00x10-11 kg each and leave the nozzle and travel horizontally toward the paper at velocity u = 16.0 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 Do = 2.30 cm where there is a uniform vertical electric field with magnitude E = 770x104 N/C Fiqure 1) 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 detemines at each nozzle position whether ink is squirted onto the paper or not Part A If a drop is to be deflected a distance d = 0.340 mm by the time it reaches the end of the deflection plate, what magnitude of charge q mlust be given to the drop? Assume that the density of the ink drop is 1000 kg/m3, and ignore the effects of gravity Express your answer numerically in coulombs. Hints Figure 1 1011 Submit My Answers Give Up Do Continue

Explanation / Answer

Here ,

m = 1 *10^-11 Kg

v = 16 m/s

E = 7.70*10^4 N/C

Do = 2.30cm = 0.023 m

A)

Here , time in field = distance/speed

Time = 0.023/16

Time = 1.43*10^-3 s

Here , Now , for d = 0.340 mm ,

Using second equation of motion ,

d = 0.5*at^2

0.340 * 10^-3 = 0.5 *a*(.00143)^2

a = 332.53 m/s^2

Now , Newton's seecond law of motion ,

ma = qE

332.53* 1 *10^-11 = q* 7.70 *10^4

solving for q

q = 4.31* 10^-14 C

the charge on the drop is 4.31 * 10^-14 C

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