Charge deflection in ink jet printer

Click For Summary
SUMMARY

The discussion focuses on calculating the charge required for ink drops in an inkjet printer to achieve a specific deflection. The drops, with a radius of 15 μm and mass of 1.41 x 10-11 kg, travel at 20 m/s through a uniform electric field of 9.1 x 104 N/C. To deflect the drops by 0.30 mm over a distance of 2.0 cm, the necessary charge was calculated to be approximately 5.268 picoCoulombs. Participants confirmed the methodology and results, discussing the application of fundamental physics equations.

PREREQUISITES
  • Understanding of electrostatics, specifically electric fields and forces.
  • Familiarity with kinematics, including motion equations.
  • Knowledge of charge calculations in electric fields.
  • Basic proficiency in physics problem-solving techniques.
NEXT STEPS
  • Study the relationship between force and electric fields using F=qE.
  • Explore kinematic equations for projectile motion in physics.
  • Learn about the properties and behavior of charged particles in electric fields.
  • Investigate practical applications of electrostatics in inkjet printing technology.
USEFUL FOR

This discussion is beneficial for physics students, engineers working with electrostatic devices, and professionals involved in inkjet printer technology and design.

steve1234
Messages
3
Reaction score
0

Homework Statement



In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The pattern on the paper is controlled by an electrostatic valve that determines at each nozzle position whether ink is squirted onto the paper or not. The ink drops, 15 μm in radius, have a mass of 1.41 x 10-11 KG leave the nozzle and travel toward the paper at 20 m/s. The drops pass through a charging unit that gives each drop a positive charge q when the drop loses some electrons. The drops then pass between parallel deflecting plates 2.0 cm in length where there is a uniform vertical electric field with magnitude 9.1 x 104 N/C. If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plate, what magnitude of charge must be given to the drop? (Answer in pC -- pico Coulombs.)


Homework Equations



F = (k*q1*q2)/r^2 , where k=8.99e9
E(Electric field) = F/q

The Attempt at a Solution



I'm pretty stuck on this one and any help on how to do this or to get me started would be appreciated.
 
Physics news on Phys.org
What is the force exerted on a charge in an electric field? Your second equation should give you the clue.
 
F=qE?
how would this help you to find this force when we are missing two variables?
 
first we found time t = d/v t=.001s

X = Xo + volt + .5at^2 to find our a

F=ma=qE and solve for q, q=ma/E

final answer came out to be 5.268 picoCoulombs, does that sound correct?
 
hey did that answer work for you?
 
Sounds ok but I get a different answer. What did you get for a?
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
11K
Replies
6
Views
9K
  • · Replies 12 ·
Replies
12
Views
16K