Operation of Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper 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 have a radius of 15.0 nano meters and leave the nozzle and travel toward the paper at a velocity of 25.0 meter/sec. The drops pass through from a a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops hen pass between parallel deflecting plates of length 1.60 centimeters where there is a uniform vertical electric field with a magnitude of 8.10x10 4th neuton/coulumbs. If a drop is to be deflected a distance of 0.290 milimeters by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is 1000 kilo grams/meter cubed)

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Originally Posted by cjssg1

Operation of Inkjet Printer. In an inkjet printer, letters are built up by squirting drops of ink at the paper 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 have a radius of 15.0 nano meters and leave the nozzle and travel toward the paper at a velocity of 25.0 meter/sec. The drops pass through from a a charging unit that gives each drop a positive charge q by causing it to lose some electrons. The drops hen pass between parallel deflecting plates of length 1.60 centimeters where there is a uniform vertical electric field with a magnitude of 8.10x10 4th neuton/coulumbs. If a drop is to be deflected a distance of 0.290 milimeters by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Assume that the density of the ink drop is 1000 kilo grams/meter cubed)

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Time spent by the droplet inside the plates is t=L/V, where L is the length of the plates, and V - droplet's speed parallel to the plates. Let's call the axis parallel to the plates X, and the axis parallel to the electric field - Y. Obviously, the speed of the droplet along the axis X does not change, as there is no force in this direction. The speed along Y changes due to the force q*E, which is constant and acts on the droplet while it is inside the plates. This force adds an extra component to the droplets speed through constant acceleration ( a=(q*E)/m, here a is acceleration, m - mass of the droplet (mass=density*volume) ), which leads to the displacement. This displacement is due to accelerated motion, therefore total displacment D=(1/2)*a*(t^2). Combination all three equation gives the fourth one, which can be solved for q. The only question here is whether to take the gravitational force into account. The answer is no due to two reasons. The first is that we do not know relative orientation of the electric field and gravitation (they either coinside or opposite to each other). The second is that probably the electric field is so much stronger than the gravity, that we can omit the gravity and its effect on the droplet's speed.