In an inkjet printer, letters are built up by squirting drops of ink at the pape
ID: 2049091 • Letter: I
Question
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 have a radius of 12.0 mu { m m} and leave the nozzle and travel toward the paper at a velocity of 23.0 { m 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 1.60 { m cm} where there is a uniform vertical electric field with a magnitude of 7.70×104 { m N/C}.If a drop is to be deflected a distance of 0.340 { m mm} 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;{ m kg/m^3}).
Solve for Q units should be in Coulombs
Explanation / Answer
q= -1.63 * 10 ^-16 C ///////////////////… This is how I got this answer (It might be possible that I made a mistake in the calculation but the general solution procedure is the same, You can re-do the sums to make sure they are correct) ======================================… The data we have : ============== Radius of drop= 20 micro meter Density of ink = 1000 kg/m^3 Mass of a drop =? Volume of sphere =4/3*Pi*r^3 Density=m/v >>> m = 1.06* 10^(-11) kg ----------------------------- E(electric field)=8.35*10^4 N/C v (velocity of drop) =23 m/s y( deflection on the paper ) =0.25 mm L(length of plates) =2.05 cm ------------------------- Unknown factor : q (electric charge of drop) =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-… Suppose the time it takes for the drop to travel through the plates is t Thus we have : ## y=1/2at^2 ## and ## L=v t ## These are simple formulas for the movement of any object -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.… We also know from newton's second law : ## F= ma ## The forces that act on the drop are its weight and the force of electric field thus : ## -mg+q(-E)=ma ## Since the drop is very small it's weight is almost nothing compared to the electric field force thus we can omit it. ## -qE = ma ## Now we combine the 3 equations ## y= 1/2 (-qE/m) (L/v)^2 ## >>> ## q= - (2ymv^2) / EL^2 ## -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.… Now all we need to do is to replace the known factors with given quantities and calculate the q ## q = {-2(0.25*10^-4) (1.06*10^-11)(23^2)} / {(8.35*10^4)(2.05*10^-2)} ## ############ q= 1.63*10^-16 C ###########
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