**Operation Of An Inkjet Printer Mastering Physics**

Help with physical problems with ink jet printers? 3

Context Information:

In an inkjet printer, letters and images are created by drawing drops of ink onto a sheet of paper with a fast-moving nozzle. The pattern on the paper is controlled by an electrostatic valve which determines the position of each nozzle whether the ink has been sprayed on the paper or not.

Question:

Each ink drop is m m = 1.00 * 10 (11) kg and exits the nozzle and travels towards the paper in longitudinal direction at a speed of v = 25.0 m / s. The P drop unit passes through the charge, giving each drop a positive charge with a loss of one electron. The length of the drop p between the parallel baffle plates is D = 2.40 cm, with a uniform vertical electric field of size E = 8.5 * 10 4 N / C. When a drop reaches the end distance d = 0.320 mm. What is the amount of fog charge q applied to each drop of deflector? Imagine that the density of a drop of ink is 1000 kg / m 3, ignoring the effects of gravity.

F is tested by a charged particle q in a uniform E-shaped electric field (the type of field between the charged plates in this question) is:

F = q * E

According to Newton's second law, F = m * a, so if a particle has m m, it experiences constant velocity:

a = q * E / m

The equation of motion for a particle with constant velocity is:

Delta = 0.5 * a * t 2 = (q * E / (2 * m)) * t 2

(2 * m * delay) / (E * t 2) = q

Where the delta is changed from the sung position to the t time position.

No force acts when it falls in the direction of its initial velocity (parallel to the plate), so the velocity of the parallel fall of the plate is constant. This means that it is important to:

t = 2.4 cm / (25 m / sec) = 9.6 * 10 4 sec

So that the drops pass through the plate. This is the moment when the fall accelerates the background, so:

2 * (1.00 * 10 11 kg) * (3.2 * 10 4 m) / ((8.5 * 10 4 N / A) * (9.6 * 10 4 sec) 2) = q

8.17 * 10 14 C = q

This is how inkjet printers work.

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D:

Help with physical problems with ink jet printers?

Context Information:

In an inkjet printer, letters and images are created by spraying ink drops on a sheet of paper with a fast moving nozzle. The pattern on the paper is controlled by an electrostatic valve which, at the position of each nozzle, determines whether the ink ...

The best answer can be found on this page >> m = 1.06 * 10 (11) kg E (electric field) = 8.35 * 10 4 N / C v (falling speed) = 23 m / s (paper deviation) = 0.25 mm L (panel length) = 2.05 cm Unknown element: q (reduction of electric charge) ====================== ... board The time to fall is t so we have: ## y = 1 / 2at 2 ## y ## L = vt ## This is a simple formula to move anything ... . . . . . ....... We also know Newton's second law: ## F = ma ## The force acting on a falling electric field is as follows: ## mg + q (E) = ma # # Since the drop is very small, its weight is almost zero compared to the strength of the electric field, so we can eliminate it. ## qE = ma ## Now we come to 3 equations ## y = 1/2 (qE / m) (L / v) 2 ## >>> ## q = (2ymv 2) / EL 2 # # ........................ We just need to change the kno factor to the given size and q ## q = {2 (0.25) Have to calculate * 10 4) (1.06 * 10 11) (23 2)} / {(8.35 * 10 4) (2.05 * 10 2)} ## ## ###### ## ## = 1.63 * 10 16 C ############