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

1.2×10⁻⁸ g . The drops will leave the nozzle and travel toward the paper at 50 m/s, 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 cm long, where there is a uniform vertical electric field with magnitude 8.0×10⁴ N/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 0.35 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

B) How many electrons must be removed from the drop to give it this charge?

C)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 0.35-mm deflection?

Answer 1

A)

consider the motion in horizontal direction:

V_ox = velocity = 50 m/s

X = distance travelled = length of plates = 2 cm = 0.02 m

t = time taken

Using the equation

X = V_ox t                                           since there is no acceleration along the x-direction

0.02 = 50 t

t = 0.0004 sec

consider the motion in Y-direction

V_oy = initial velocity = 0 m/s

a = acceleration = ?

Y = deflection = 0.35 x 10^-3 m

t = time = 0.0004 sec

using the equation

Y = V_oy t + (0.5) a t²

0.35 x 10^-3 = (0) (0.0004) + (0.5) a (0.0004)²

a = 4375 m/s²

acceleration caused to electric field is given as

a = qE/m

4375 = q (80000)/(1.2 x 10^-11)

q = 6.6 x 10^-13 C

b)

n = number of electrons removed

e = magnitude of charge on each electron = 1.6 x 10^-19 C

using the equation

q = ne

n = q/e = (6.6 x 10^-13)/(1.6 x 10^-19)

n = 4.125 x 10⁶

c)

consider the motion in horizontal direction:

V_ox = velocity = 25 m/s

X = distance travelled = length of plates = 2 cm = 0.02 m

t = time taken

Using the equation

X = V_ox t                                           since there is no acceleration along the x-direction

0.02 = 25 t

t = 0.0008 sec

consider the motion in Y-direction

V_oy = initial velocity = 0 m/s

a = acceleration = ?

Y = deflection = 0.35 x 10^-3 m

t = time = 0.0008 sec

using the equation

Y = V_oy t + (0.5) a t²

0.35 x 10^-3 = (0) (0.0008) + (0.5) a (0.0008)²

a = 1093.75 m/s²

acceleration caused to electric field is given as

a = qE/m

1093.75 = q (80000)/(1.2 x 10^-11)

q = 1.6 x 10^-13 C