Electric field

[匿名]

2009-03-05 2:04 pm

The deflection plates of an ink jet printer are set up horizontal and parallel one above the other; both are square, of side 3.0cm. They are charged such that the electric field in the region between them is uniform, pointing vertically downwards, and of magnitude 8.0x10^6N/C. An ink drop of mass 4.0 x 10^-10kg and charge -1.3 x 10^-13C enters the region between the plates, travelling horizontally at 30m/s parallel to a side of the plates.

(1) Draw a diagram to represent the situation and label the axis reasonably.

(2) Calculate the force acting on the ink drop due to the electric field.

(3) How long does the ink drop travel the length of the plates?

(4) Through what vertical displacement is the drop deflected at the end of its passage between the plates?

回答 (1)

Prof. Physics

2009-03-05 3:57 pm

✔ 最佳答案

a. Please visit the following link for your diagram:

http://i726.photobucket.com/albums/ww265/physicsworld2010/physicsworld01Mar050749.jpg

b. Force acting on the ink drop due to the electric field, F

= qE

= (1.3 X 10^-13)(8.0 X 10^6)

= 1.04 X 10^-6 N

(Since the electric field is pointing vertically downwards, so the upper plate must be positively charged and the lower one is negatively charged.)

c. The horizontal velocity of the ink drop is not affected. The ink is moving with uniform horizontal speed.

So, by Time = Distance / Speed

Time = 0.03 / 30

Time, t = 1 X 10^-3 s

d. The net upward force acting on the ink drop, F'

= F - mg

= (1.04 X 10^-6) - (4.0 X 10^-10)(10)

= 1.036 X 10^-6 N

By Newton's 2nd law of motion,

F' = ma

1.036 X 10^-6 = (4.0 X 10^-10)a

Upward acceleration, a = 2590 ms^-2

By equation of motion, s = ut + 1/2 at^2

s = 0 + 1/2 (2590)(1 X 10^-3)^2

Upward displacement, s = 1.295 X 10^-3 m = 0.1295 cm

參考: Physics king

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