Electrostatic Printer and two plate capacitors deflecting charged ink droplets

[Dester]

Homework Statement

One type of ink-jet printer, called an electrostatic ink-jet printer, forms letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms 30-μm-diameter drops of ink, charged them by spraying 800,000 electrons on the surface, and shoots them toward the page at a speed of 20 m/s. Along the way, the drops pass through two parallel electrodes that are 6.0 mm long, 4.0 mm wide, and spaced 1.0 mm apart. The distance from the center of the plates to the paper is 2.0 cm. To form the letters which have a maximum hieght of 6.0 mm, the drops need to be deflected up or down at maximum if 3.0 mm. Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg.m^3

What amount of charge is needed on each electrode to produce this electric field

Homework Equations

The Attempt at a Solution

I know qE/m, I found m and q and also E via part A I solved, but now I'm stumped, can anyone give me hand?

 

[haruspex]

How long will a drop take to reach the paper? What vertical velocity will it need to achieve the desired deflection? How long does it have to acquire that velocity? What acceleration does that imply?

 

[Basic_Physics]

Doesn't it follow a parabolic trajectory between the plates, sort of acting like a stronger gravitational field?

 

[haruspex]

Doesn't it follow a parabolic trajectory between the plates, sort of acting like a stronger gravitational field?

Sure, but most of the deflection (in terms of distance, as opposed to angle) will occur across the 2cm gap to the paper.

 

[Nickie]

To calculate the amount of charge needed on each electrode, we can use the equation q = mE/g, where q is the charge, m is the mass of the ink droplet, E is the electric field, and g is the gravitational force.

First, we need to calculate the mass of the ink droplet. We know that the diameter of the droplet is 30 μm, and its density is 800 kg/m^3. Using the formula for volume of a sphere (V = (4/3)πr^3), we can find the volume of the droplet to be approximately 14.1 x 10^-12 m^3. Then, using the formula for density (ρ = m/V), we can calculate the mass of the droplet to be 11.3 x 10^-9 kg.

Next, we need to calculate the electric field. We can use the equation E = V/d, where V is the potential difference between the plates and d is the distance between them. We know that V = 800,000 electrons x 1.6 x 10^-19 C/electron = 0.128 C, and d = 1.0 mm = 0.001 m. Plugging these values into the equation, we get E = 128 N/C.

Finally, we can calculate the amount of charge needed on each electrode using the formula q = mE/g. Plugging in the values we have calculated, we get q = (11.3 x 10^-9 kg)(128 N/C)/9.8 m/s^2 = 1.46 x 10^-6 C.

Therefore, we need a charge of approximately 1.46 microcoulombs on each electrode to produce the electric field necessary for the ink droplets to be deflected by a maximum of 3.0 mm.