How Much Charge is Needed for an Ink Drop to Deflect in an Inkjet Printer?

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    Ink Jet Physics
AI Thread Summary
The discussion focuses on calculating the charge needed for an ink drop in an inkjet printer to achieve a specific deflection while traveling through an electric field. The ink drops, which are positively charged, are influenced by both electric and gravitational forces as they move towards the paper. Key equations involve time of travel, distance, and forces acting on the drop, including the relationship between mass, acceleration, and charge. The challenge lies in determining the time taken for the drop to traverse the distance and deriving an equation for the required charge. The conversation highlights the complexity of balancing forces to achieve the desired deflection.
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Homework Statement



In the somewhat ‘archaic’ inkjet printers, 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, αr, in radius (here, α is a positive, dimensionless constant), leave the nozzle and travel toward the paper at v 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 pass between parallel deflecting plates d cm in length where there is a uniform vertical electric field with magnitude E N/C. If a drop is to be deflected β mm (here β is a positive number) 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 ρ kg/m3.)

Homework Equations



1) t = D0 / v

2) d = ( a * t^2 ) / 2

3) F = m * a

4) F = E * Q

The Attempt at a Solution



Having a lot of trouble...
 
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Draw a triangle with a length d and a height beta. The x velocity is constant in this case, and there is an electric force acting on the drop in an upward direction and a gravitational force in the downward direction so

Fy = QE - mg = may

and over a distance d it has moved up a distance beta

You need to figure out how long it takes for the drop to travel a distance d and figure out an equation for the acceleration upwards.

Then solve it in terms of q
 
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