In an inkjet printer, letters are built up by squirting drops of ink horizontally at a sheet of paper from a rapidly moving nozzle.

ankarskogC

ankarskogC

Answered question

2020-11-08

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In an inkjet printer, letters are built up 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 thepaper or not. The ink drops have a radius 15.0μm and leave the nozzle and travel horizontally toward the paper at speed v=18.0 m/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 D0=2.45 cm , where there is a uniform vertical electric field with magnitude E=8.40×104 N/C
If a drop is to be deflected a distance d= 0.270 mm by the time it reaches the end of the deflection plate,what magnitude of charge q must be given to the drop? (Assume that the density of the ink drops 1000 kgm3, and ignore the effects of gravity.)

Answer & Explanation

irwchh

irwchh

Skilled2020-11-09Added 102 answers

The only challenging aspect of this problem is sorting out and using all the information that was provided to you. It sounds complicated.
The horizontal motion is at constant velocity. They tell you the drops have constant horiz velocity of 18 m/s and they must travel a horizontal distance of 2.45 cm, or 0.0245m. So what can you get from this? With distance and speed,you can determine the amount of time it takes for the drops to get from one end to the other: 
time = dist / speed =0.0245 m / 18 m/s = 0.001361 sec 
Now consider the vertical motion. There is no initial vertical velocity. But there is a constant force from the Field: F=qE which causes a constant acceleration. Using basic kinematics, you can relate the distance, acceleration and time by: 
d=(12)at 
you are given the vertical distance is 0.270 mm or 0.000270m 
You can therefore refer to the force of the E field's force as the acceleration the drop must experience:
a=2dt 
=20.000270 m0.001361 s 
=291.5 m/s 
You can now relate the force and acceleration acting on the drop:
F=ma or qE=ma you have values for E and a, you are asked to find q. Do you need a value for m, the mass of the drop. You are given its size(radius) and density. 
So mass=densityvolume=density(43) 
To make things a little easier, we will use 1 gram/ cc for density, which is the same as 1000 kg/m 
(just different units) and we will express r in cm. 
So: m=1 gram / cc*(4/3) 
=1.4137×10 
Now: qE=ma or q=maE=1.4137×10 
kg291.5 ms=8.40×10 
NC=4.906×10

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