Electric field strength in an ink-jet printer

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1. Feb 22, 2017

pbj_sweg

1. The problem statement, all variables and given/known data
One type of ink-jet printer, called an electrostatic ink-jet printer, forms the 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 29.0 μm diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page with a horizontal velocity of 16.0 m/s. Along the way, the drops pass through the long axis of two horizontal, 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 electrodes to the paper is 2.80 cm. To form the letters, which have a maximum height of 6.0 mm, the drops need to be deflected up or down a maximum of 3.0 mm. Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg/m3.

What electric field strength is needed between the electrodes to achieve this deflection?

2. Relevant equations
$$y_f = y_i+v_{0y}t+\frac{1}{2}a_yt^2$$
$$\vec{F}_{net}=m\vec{a}$$
$$F_E=qE$$
$$d = v_0t$$
$$\rho = \frac{m}{V}$$
$\Delta{x}$ = 6.0 mm, $\Delta{y}$ = 3.0 mm

3. The attempt at a solution
(1) Found time that a drop of ink is inside the electrode by $d=v_0t$.
$t = \frac{\Delta{x}}{v_0}$ = 0.000375 sec

(2) Found the acceleration in the y-direction with $y_f = y_i+v_{0y}t+\frac{1}{2}a_yt^2$
$a_y = \frac{2\Delta{y}}{t^2}$ = 42666.7 m/s^2 (I was a bit hesitant to continue with this number as it seemed too large to be correct).

(3) Found charge on the drop of ink (e is the charge of an electron): $800000\times{e} = 1.28176\times{10^{-13}}$

(4) $\vec{F}_{net} = \vec{F}_E+\vec{F}_G = m\vec{a}$
As there is no force in the x direction, we can solve for forces in y-direction:
$F_y = F_E-F_G = ma_y$
$qE-mg = ma_y$
$E = \frac{ma_y+mg}{q}$
$E = 3.4\times10^6 \frac{N}{C}$

I'd appreciate it if someone could guide me to the proper solution as my answer is incorrect.

Sidenote - the mass of the drop was found with the following calculation:

Volume of drop = $V = \frac{4}{3}\pi{r^3}$ and radius = $r = \frac{29}{2} \mu{m} = 1.45\times10^{-5} m$
$V = \frac{4}{3}\pi{r^3} = 1.277\times10^{-14}$
$m=\rho{V} = 800V = 1.0216\times10^{-11} kg$

Last edited: Feb 22, 2017
2. Feb 22, 2017

Staff: Mentor

Hi pbj_sweg,

I don't think that you should assume that the ink drop trajectory just grazes the edge of the plate. That is, the deflection of the ink does not have to be 0.5 mm (your Δy/2). Instead, I think you should make the Δy at the end of the plates an unknown.

Start by assuming that there is some electric field E and write an expression for the acceleration of the ink drop using it.