The ink drops have a mass m = 1.00x10-11 kg each and leave the nozzle and travel

Question

The ink drops have a mass m = 1.00x10-11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 25.0 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 D0 = 1.50cm , where there is a uniform vertical electric field with magnitude E = 8.50x104 N/C . (Figure 1) If a drop is to be deflected a distance d = 0.310 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 drop is 1000kg/m3 , and ignore the effects of gravity. Express your answer numerically in coulombs.

Explanation / Answer

The force, F, experienced by a particle with charge q in a uniform electric field of magniqude E (the type of field between the charged plates in this question) is: F = q*E By Newton's second law, F = m*a, so if the particle has mass m, it will undergo a constant acceleration of: a = q*E/m The equation of motion for a particle undergoing constant acceleration is: delta-y = 0.5*a*t^2 = (q*E/(2*m))*t^2 (2*m*delta-y)/(E*t^2) = q where delta-y is the change in position at time t, from the starting position. There are no forces acting on the droplet in the direction of its initial velocity (parallel to the plates), so the droplet's velocity parallel to the plates is constant. That means it will take: t = 2.4cm/(25 m/s) = 9.6*10^-4 sec for the droplet to traverse the plates. This is the amount of time that the droplet is subjected to the transverse acceleration, so: 2*(1.00*10^-11 kg)*(3.2*10^-4 m)/((8.5*10^4 N/C)*(9.6*10^-4 sec)^2) = q 8.17*10^-14 C = q

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