You've been hired to design the hardware for an ink jet printer. You know that these printers use a deflecting electrode to cause charged ink drops to form letters on a page. The basic mechanism is that uniform ink drops of about 30 microns radius are charged to varying amounts after being sprayed out towards the page at a speed of about 20 m/s. Along the way to the page, they pass into a region between two deflecting plates that are 1.6 cm long. The deflecting plates are 1.0 mm apart and charged to 1500 volts. You measure the distance from the edge of the plates to the paper and find that it is one-half inch. Assuming an uncharged droplet forms the bottom of the letter, how much charge is needed on the droplet to form the top of a letter 3 mm high (11 pt. type)

the required charged is 7.06 × 10⁻¹³ C

Explanation:

Given that;

Radius = 30 microns = 30 × 10⁻⁶

Speed v = 20 m/s

length x = 1.6 cm = 0.016 m

spacing d = 1.0 mm = 0.001 m

Voltage V = 1500 V

from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.

Electric field E = V/d

E = 1500 V /  0.001 m

E = 1.5 × 10⁶ V/m

Mass of ink drop m = pv

m = 10³ kg/m³ × πr³

m = 1000 kg/m³ × π × (30 × 10⁻⁶)³

m = 1.131 × 10⁻¹⁰ Kg

Time taken to travel t =  x / sped

t = 0.016 m / 20 m/s

t = 0.0008 s

From the kinematic equation

to form the top of a letter 3 mm ( 0.003 m )high

y = at²

2y = at²

a = 2y/t²

we substitute

a = (2 × 0.003 m) / (0.0008 s)²

a =  9375 m/s²

Now Force F = Eq = ma

so

q = ma / E

we substitute

q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )

q = 7.06 × 10⁻¹³ C

Therefore, the required charged is 7.06 × 10⁻¹³ C

e/c, where e is its total energy

electric field is a function of 1/r^2

yes, the force per unit charge for a point charge is inverse square law.