In an inkjet printer, tiny drops of ink of inertia m are given a charge q and th
ID: 2000222 • Letter: I
Question
In an inkjet printer, tiny drops of ink of inertia m are given a charge q and then fired toward the paper at speed v. They first pass between two charged plates of length that create a uniform electric field of magnitude E between them (Figure 1) . The field direction is perpendicular to the plates and to the initial path of the drops. The electric field deflects each drop from its initial horizontal path and is used to control what is printed on the paper.
a.
Derive an expression for how far in the vertical direction a drop is deflected as it passes between the plates. Use the notation l for the length of the plates .
Express your answer in terms of the variables m, q, v, l, E, and acceleration due to gravity g.
b.For a drop inertia of 2.0×1010 kg , an electric field strength of 1.2×106 N/C , a plate length of 13 mm , and a drop speed of 27 m/s , calculate the drop charge necessary to result in a deflection of 1.3 mm.
Explanation / Answer
(a) An expression for distance in the vertical direction, a drop deflected as it passes between the plates which is given as :
from newton's second law, we have
F = m ay { eq.1 }
we know that, F = q E
q E = m ay
ay = q E / m { eq.2 }
Let, t be the time required for the drop to pass the region and deflect.
using equation of motion 2, we have
y = v0 t + (1/2) ay t2
where, v0 = initial speed = 0
y = (1/2) ay t2 { eq.3 }
where, t = L / v
then, we get
y = q E L2 / 2 m v2
Answer in terms of m, q, v, L, and E.
(b) The drop charge necessary to result in a deflection of 1.3 mm which is given as :
using part-a equation, y = q E L2 / 2 m v2
(1.3 x 10-3 m) = q (1.2 x 106 N/C) (13 x 10-3 m)2 / 2 (2 x 10-10 kg) (27 m/s)2
(3790.8 x 10-13 N) = (202.8 N/C) q
q = (3790.8 x 10-13 N) / (202.8 N/C)
q = 18.7 x 10-13 C
q = 1.87 x 10-12 C
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