In an inkjet printer, letters are built up by squirting drops of ink at the pape

Question

In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of 1.6 x 10^-8 g each, leave the nozzle and travel toward the paper at 21 m/s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 cm long where there is a uniform vertical electric field with magnitude 7.5 x10^4 N/c. If a drop is to be deflected 0.35 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? A particle with charge + 8.30 nC is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right, After it has moved 7.00 cm, the additional force has done 6.30 x 10^-5 J of work and the particle has 4.35 x 10^-5 J of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

Explanation / Answer

The drop is deflected due to electric force

F = q*E
m*a = q*E

accelaration a = q*E/m

apply the equations from kinematics

y = 0.5*a*t^2

t be the time required for the drop to pass the region and deflect

t = l/v

l = 2 cm = 0.02 m

v = 21 m/s

t = 0.02/21 =9.52*10^-4 S

then y = 0.5*(q*E/m)*t^2

0.35*10^-3 = 0.5*q*7.5*10^4*(9.52*10^-4)^2/(1.6*10^-11)

charge q = 1.647 *10^-13 C = 0.1647*10^-12 C

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Work done by electric force + Work done by the another force = change in kinetic energy

Work done by electric force = (4.35*10^-5)-(6.3*10^-5) = -1.95*10^-5 J

since work done is scalar.Hence the work done by the electric force = W= 1.95*10^-5 J

B) W = q*V = 1.95*10^-5

potential difference is V = (1.95*10^-5)/q =(1.95*10^-5)/(8.3*10^-9)

V = 2349.4 V

C) magnitude of electric field is E = V/d = 2349.4/0.07 = 3.35*10^4 V/m

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