A charged drop of ink (which may be considered as a proton, so its mass is that

Question

A charged drop of ink (which may be considered as a proton, so its mass is that of a proton) is projected between the deflecting plates in an inkjet printer, travelling to the right at a speed of 1.276 10-5 m/s, as shown in the figure. The distance between the two plates is 1.64 cm. The ink drop enters the plates halfway between the top plate and the bottom plate; that is, a distance r = 0.820 cm from each plate, as shown in the figure. The plates have a 2.70 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the ink drop traverse before it hits the bottom plate?

Explanation / Answer

given

vox = 1.276*10^5 m/s

E = 2.7*10^4 N/c

in between the plates,

Fe = q*E

m*ay = q*E

ay = q*E/m

= 1.6*10^-19*2.7*10^4/(1.67*10^-27)

= 2.59*10^12 m/s^2

let t is the time taken for the ink drop to hit the bottom plate.

use, y = voy*t + (1/2)*ay*t^2

y = 0 + (1/2)*ay*t^2

==> t = sqrt(2*y/ay)

= sqrt(2*0.0082/(2.59*10^12))

= 7.96*10^-8 s

horizontal distance travelled before hitting the bottom plate, x = vox*t

= 1.276*10^5*7.96*10^-8

= 0.0101 m or 1.01 cm <<<<<<<-----------Answer

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