Question

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.5×10−8g. The drops will leave the nozzle and travel toward the paper at 50 m/s in a horizontal direction, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude 8.0×104N/C. Your team is working on the design of the charging unit that places the charge on the drops.

(a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge?
(b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 m/s, what q value is needed to achieve the same 0.30-mm deflection?

Answer 1

A) q = 7.03 × 10^(-13) C

n ≈ 4388265 electrons

B) q = 1757.8 × 10^(-16) C

n ≈ 1097253 electrons

Step-by-step explanation:

We are given;

Mass; m = 1.5 × 10^(−8)g = 1.5 × 10^(-11) kg

Speed; v = 50 m/s

Electric Field; E = 8 × 10⁴ N/C

Distance between plates; s = 2 cm = 0.02 m

A) Now, speed = distance/time

So, time(t) = distance/speed = 0.02/50 = 0.0004 s

From Newton's first law of motion, we know that;

d = ut + ½at²

But u is initial velocity and in this case it's zero.

But we are told that a drop is to be deflected by 0.30 mm. So, d = 0.3 × 10^(-3) m

Thus;

0.3 × 10^(-3) = 0 + ½a(0.0004)²

a = 3750 m/s²

Now, we know that force in motion normally can be expressed as;

F = ma

But in electric field, it's;

F = qE

Thus;

qE = ma

So, charge is; q = ma/E

Plugging in the relevant values;

q = (1.5 × 10^(−11) × 3750)/(8 × 10⁴)

q = 7.03 × 10^(-13) C

Now, number of electrons is given by the formula;

n = q/e

Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron

So; n = (7.03 × 10^(-13))/(1.602 × 10^(-19))

n ≈ 4388265 electrons

B) We are told speed is now 25 m/s.

Thus;

time(t) = distance/speed = 0.02/25 = 0.0008 s

From Newton's first law of motion, we know that;

d = ut + ½at²

But u is initial velocity and in this case it's zero.

d remains 0.3 × 10^(-3) m

Thus;

0.3 × 10^(-3) = 0 + ½a(0.0008)²

a = 937.5 m/s²

Now, we know that force in motion normally can be expressed as;

F = ma

But in electric field, it's;

F = qE

Thus;

qE = ma

So, charge is; q = ma/E

Plugging in the relevant values;

q = (1.5 × 10^(−11) × 937.5)/(8 × 10⁴)

q = 1757.8 × 10^(-16) C

Now, number of electrons is given by the formula;

n = q/e

Where e is charge on electron with a value of 1.602 × 10^(-19) C/electron

So; n = (1757.8 × 10^(-16) )/(1.602 × 10^(-19))

n ≈ 1097253 electrons