Problem 1. In an inkjet printer, tiny drops of ink of inertia m are given a char

Question

Problem 1. 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 l that create a (approximately) uniform electric field of magnitude E bctween them. 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) Show that the amount of vertical deflection is given by Ay (qE/m)(l/v)2 (ignore gravity). (b) For a drop of inertia 1.5 x 10-10 kg, an electric field strength of 1.2 x 106 N/C, a plate of length 10 mm, and a drop speed of 20 m/s, calculate the drop charge necessary to result in a deflection of 1.3 mm. (c) What will be the drop's speed as it leaves the region between the plates (d Is it acceptable to ignore gravity? What is the ratio of the electric force to the gravitational force on the drop?

Explanation / Answer

a)

Force acting on the drops :

F = Eq

So, vertical acceleration ,a = F/m = Eq/m

So, the distance travled in vertical direction, s = ut + 0.5*at^2

for this u = 0 , t = time taken

Now, in horizontal direction : t = horizontal distance / horizontal speed = l/v

So, s = 0 + 0.5*(Eq/m)*(l/v)^2 = (Eq/m)*(l/v)^2/2 <----- PROVED

b)

Ay = 1.3 mm = 0.0013 m

m = 1.5*10^-10 kg

E = 1.2*10^6 N/C

l = 10 mm = 0.01 m

v = 20 m/s

So, Ay = 0.0013 = (1.2*10^6*q/(1.5*10^-10))*(0.01/20)^2/2

So, q = 1.3*10^-12 C

c)

speed in horizontal = 20 m/s

speed in vertical , v = u +at = 0 + (Eq/m)*t

= (Eq/m)*l(/v) <---- Plug in the values to get answer if you want

e)

Ratio of Electric force / gravity = Eq / mg = 1.2*10^6*1.3*10^-12 / (1.5*10^-15*9.8)

= 1.06*10^8

So, electric force is very much greater than gravity. So, gravity can be ignored

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