In an inkjet printer, letters are built up by squirting drops of ink at the pape

Question

In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of 1.3 Times 10^-8 g each, leave the nozzle and travel toward the paper at 19 m/s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 cm long where there is a uniform vertical electric field with magnitude 8.5 Times 10^4 N/C. If a drop is to be deflected 0.26 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

Explanation / Answer

Assume that the density of the ink drop is 1000 kg/m3, and ignore the effects of gravity.
Remember that the force experienced by the drop is equal to:    F = qE

Since F = ma, we also know that:          ma = qE ======> a = qE/m

Now we can use the acceleration to find the change in position (deflection):

x = 1/2at2
x = 1/2 * (qE/m) * t2

And we can use this to find q:

x = 1/2 * (qE/m) * t2
2x / (t2) = qE/m
q = (2 * x * m) / (E * t2)

Since the plates are 2.0 cm long, the time will be:

t = 0.02 m / 19 m/s = 0.0010526 s

Therefore:

q = (2 * d * m) / (E * t2)
q = (2 * 0.00026 * (1.3*10^-11)) / ((8.5*10^4) * (0.0010526^2))
q = 7.18*10^-14 C

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.