Answer :
The magnitude of charge q that must be given to drop is;
q = 6.36 * 10⁻¹⁴ N/C
Electric field calculations
We are given;
Deflection distance; d = 0.35 mm = 0.00035 m
Density; ρ = 1000 kg/m³
Speed; v = 18.0 m/s
horizontal distance; D₀ = 2.45 cm = 0.0245m
Electric field; E = 8.40 × 10⁴ N/C.
Thus;
Time = distance/speed
Time = 0.0245/18
time; t = 0.001361 sec
We know from newtons second equation of motion that;
d = u + ¹/₂at²
where;
u is initial velocity
a is acceleration
t is time
Initial velocity here is zero.
Thus;
d = ¹/₂at²
Making acceleration (a) the subject gives;
a = 2d/t²
a = (2 * 0.00035)/(0.001361²)
a = 377.9 m/s²
Formula for density is;
ρ = m/V
m = ρV
For volume, we will use ⁴/₃πr³ m³
Thus;
m = 1000 * ⁴/₃ * π * (15 * 10⁻⁶)³
m = 1.414 * 10⁻¹¹ kg
formula for the magnitude of the charge is;
q = ma/E
q = (1.414 * 10⁻¹¹ * 377.9)/(8.40 × 10⁴)
q = 6.36 * 10⁻¹⁴ N/C
The first half of this question is missing and it is;
In an inkjet printer, letters are built up 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 radius 15.0μm and leave the nozzle and travel horizontally toward the paper at speed v = 18.0 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 D₀ = 2.45 cm , where there is a uniform vertical electric field with magnitude E=8.40 × 10⁴ N/C.
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