The ink drops have a mass m = 1.00×10-11kg each and leave the nozzle and travel

Question

The ink drops have a mass m = 1.00×10-11kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 24.0 m/s. The drops pass through a charging unit that gives each drop a positive charge by causing it to lose some electrons. The drops then pass between parallel deflecting plates of length Do = 2.25 cm , where there is a uniform vertical electric field with magnitude E = 7.85×104 N/C .<br />If a drop is to be deflected a distance = 0.300 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 , and ignore the effects of gravity.

Explanation / Answer

The deflection of the charge in the electric field of length Do is d = q E Do^2 / 2 m v^2 ==> q = 2 m v^2 d / E Do^2 = 2 * 1.0 x 10^-11 * (24)^2 * 0.300/ * 7.85 x 10^4 * (2.25 x 10^-2)^2 = 8.696 x 10^-11 C Volume = mass / density = 1.0 x 10^-11 / 1000 = 1.0 x 10^-14 m^3 Therefore volume charge density = charge / volume = 8.696 x 10^-11 / 1.0 x 10^-14 = 8.696 x 10^3 C / m^3

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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