Question

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of 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 mass m = 1.00×10⁻¹¹ kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 15.0 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 = 2.20 cm , where there is a uniform vertical electric field with magnitude E = 7.70×10⁴ N/C .

If a drop is to be deflected a distance d = 0.260 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 1000 kg/m3 , and ignore the effects of gravity.

Express your answer numerically in coulombs

Answer 1

Given,

m = 1.00×10^-11 kg ; v = 15 m/s ; D0 = 2.2 cm = 0.022 m ; E= 7.7×10⁴ N/C

A)d = 0.260 mm = 0.00026 m

Let q be the charge.

We know that, force on charge due to the presence of E field is

F = q E ; Also, F = m a

m a = q E =. a = q E/m

the deflection of electron from equation of motion will be ; x = 1/2 a t²

x = 1/2 ( q E / m ) t²

Solving this for q we get

q = 2 x m / E t²

We know that, speed = distance/time

time = t = distance/speed = D0/v = 0.022 /15 = 0.0015 sec

putting this value in the equation of q.

q = 2 x m / E t² = 2 x 0.00026 x 1.00 x 10^-11 / (7.7 x 10⁴ x 0.0015^2) = 3 x 10^-14 C

Hence, q = 3 x 10^-14 C.