F5 ELECTRICITY FIELD

Ian Langley

2011-11-16 5:12 am

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Ink droplets, each of mass 1.60X10^ -10 kg, are ejected horizontally with a speed of 20m/s. When passing through the charging chamber, they are charged to a suitable value according to the input signal which determines the type and size of the output characters.

Then the ink droplets are deflected by a pair of deflection plates to a suitable position on paper. The magnitude of the electric field between the plates is
4X10^6 N/C . When there is no input signal, ink droplets are uncharged and travel along a straight line to the gutter.

Neglecting the air resistance and the gravitational force from the Earth, find the charge carried by an ink droplet if the droplet can reach a point 2.0 mm above the gutter.

( I attempted to find the charge by setting two equations. Find how much time it needed to travel the plate and how much it raised. But the ans is 5X10^-14C and I calculated 2.778X10^-14C)

Thank you.

回答 (1)

天同

2011-11-16 7:22 am

✔ 最佳答案

Time needed to pass the plates = 0.016/20 s = 8x10^-4 s
Vertical acceleration when passing the plates = q(4 x 10^6)/(1.6x10^-10)
= (2.5x10^16)q
where q is the charge on the ink droplet
Hence, vertical displacement = (1/2).[(2.5x10^16)q].(8x10^-4)^2 = (8x10^9)q

Let a be the angle that the path of the droplet makes with the horizontal when it leaves the plates. We have,
tan(a) = [0.002 - (8x10^9)q]/0.032

Vetrical velocity of the droplet when leaving the plates
= [(2.5x10^16)q].(8x10^-4) = (2x10^13)q
hence, tan(a) = (2x10^13)q/20

Therefore, (2x10^13)q/20 = [0.002 - (8x10^9)q]/0.032
solve for q gives q = 5 x 10^-14 C

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