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?11kg each and leave the nozzle and travel horizontally toward the paper at velocity v = 18.0m/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 = 2.00cm , where there is a uniform vertical electric field with magnitude E = 7.75�104N/C
1) If a drop is to be deflected a distance d = 0.310mm 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.
q = _______ C

Answer 1

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$