Question

In an ink-jet printer, drops of ink are given a controlled amount of charge before they pass through a uniform electric field generated by two charged plates. If the electric field between the plates is 1.5x106N/C over a region 1.20cm long, and drops of mass 0.12nanograms enter the electric field region moving horizontally at 10m/s, (a) how much charge on a drop is necessary to produce a vertical deflection of 0.50mm at the end of the electric field? (b) What is the vertical deflection between the point where it leaves the plates and where it strikes the paper, if the paper is 0.50cm from the end of the charged plates? Ignore other forces such as gravity and air resistance, and assume the field between the plates is vertical and constant, and drops abruptly to zero at the edge of the plates. (a) C (b) cm

Answer 1

TIme taken by a drop to cross the electric field = Horizontal distance / horizontal speed = 0.012/10 = 0.0012 s

Let the charge on each drop be Q columbs

Force on each drop, F = Q*E = Q*1.5*10⁶ N/C

Also, F = m*a

Thus, acceleration of drop, a = F/m = Q*E/m = Q*1.5*10⁶/(0.12*10^-9) = 12.5*10¹⁵*Q m/s²

Vertical deflection, S = 0.5 mm = 0.5*10^-3 m

Using equation : S = ut + 1/2at²

Here, u = initial velocity in vertical direction = 0

Putting values we get :

0.5*10^-3 = 1/2*a*0.0012²

Thus, a = 6250/9 m/s²

Thus, 12.5*10¹⁵*Q = 6250/9

Solving we get : Q = 55.56*10^-15 C

(b) Vertical velocity at the exit of field is calculated by : v = u+at

So, v = 0 + 6250/9*0.0012 = 0.83 m/s

Distance of paper from field end = 0.5 cm = 0.005 m

Time required to cover this horizontal distance = 0.005/10 = 0.0005 s

Thus, vertical deflection achieved during this time from the point of exit = vertical speed*time = 0.83*0.0005 = 4.16*10^-4 m = 0.416 mm