You are working as an engineer designing ink-jet printers. Your boss has come ac
ID: 1579491 • Letter: Y
Question
You are working as an engineer designing ink-jet printers. Your boss has come across a printer design that involves a charged drop of ink experiencing a force exerted by 100 small, fixed charged particles. As shown in (Figure 1), the particles are arranged in a ring of radius 6.00 mm, and each carries a charge of 3.00 nC . The ink drop is released at a point 6.00 mm away from the plane of the ring and along the central axis running perpendicular to the plane of the ring. Your boss gives up on calculating the initial force exerted on the charged drop, but you take one look at the setup and realize you can do the calculation in less than a minute.
Part A
Determine the magnitude of the initial force exerted on a drop that carries a charge of 9.00 nC .
Express your answer to two significant digits and include the appropriate units.
F =
Part B
Determine the direction of the initial force exerted on a drop that carries a charge of 9.00 nC .
Determine the direction of the initial force exerted on a drop that carries a charge of 9.00 .
The force is directed horizontally to the right.
The force is directed vertically upward.
The force is directed vertically downward.
The force is directed at the angle of 45 from downward direction.
The force is directed horizontally to the left.
Part C
If the printer adds to each drop of ink a charge equal to 1.00 C per kilogram of ink mass, what is the initial acceleration of each drop?
answer to two significant digits and include the appropriate units.
a =
ink drop 6.00 mm 6.00 mm. ring of charged particlesExplanation / Answer
charge on ring, Q = (100 x 3 x10^-9)
Q = 0.3 x 10^-6 C
F = k Q a / (r^2 + a^2)^(3/2)
F= (9 x 10^9 x 0.3 x 10^-6 x 9 x 10^-9)(0.006)/(0.006^2 + 0.006^2)^(3/2)
F = 0.24 N ....Ans
(B) Direction: vertically upward
(C) mass of each drop, m = (1 kg)(9 x 10^-9)
m = 9 x 10^-9 kg
a= F./ m = 26.5 x 10^6 m/s^2
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