Charge deflection in ink jet printer

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In an inkjet printer, ink drops are charged and deflected by an electric field to create patterns on paper. The problem involves calculating the charge required for a drop to be deflected 0.30 mm while traveling through a uniform electric field of 9.1 x 10^4 N/C. The force on the drop is determined using F = qE, and the acceleration is derived from kinematic equations. After calculations, one participant arrived at a charge of 5.268 picoCoulombs, but there was some disagreement on the accuracy of this result. The discussion highlights the complexities of applying physics equations to solve for charge in an electric field scenario.
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Homework Statement



In an inkjet printer, letters are built up by squirting drops of ink at the 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, 15 μm in radius, have a mass of 1.41 x 10-11 KG leave the nozzle and travel toward the paper at 20 m/s. The drops pass through a charging unit that gives each drop a positive charge q when the drop loses some electrons. The drops then pass between parallel deflecting plates 2.0 cm in length where there is a uniform vertical electric field with magnitude 9.1 x 104 N/C. If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Answer in pC -- pico Coulombs.)


Homework Equations



F = (k*q1*q2)/r^2 , where k=8.99e9
E(Electric field) = F/q

The Attempt at a Solution



I'm pretty stuck on this one and any help on how to do this or to get me started would be appreciated.
 
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What is the force exerted on a charge in an electric field? Your second equation should give you the clue.
 
F=qE?
how would this help you to find this force when we are missing two variables?
 
first we found time t = d/v t=.001s

X = Xo + Vot + .5at^2 to find our a

F=ma=qE and solve for q, q=ma/E

final answer came out to be 5.268 picoCoulombs, does that sound correct?
 
hey did that answer work for you?
 
Sounds ok but I get a different answer. What did you get for a?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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