Question

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$30\mu m$ diameter drops of ink, charges them by spraying $800,000$ electrons on the surface, and shoots them toward the page at a speed of $20m/s$. Along the way, the drops pass through two horizontal, parallel electrodes that are $6.0mm\mathrm{long},4.0mm$ wide, and spaced $1.0mm$ apart. The distance from the center of the electrodes to the paper is $2.0cm$. To form the tallest letters, which have a height of $6.0mm$, the drops need to be deflected upward (or downward) by $3.0mm$. What electric field strength is needed between the electrodes to achieve this deflection? Ink, which consists of dye particles suspended in alcohol, has a density of $800kg/m^3$ .

Short answer

The electric field strength needed is$\mathbf{7}\mathbf{.}\mathbf{06}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}\mathbf{}$

Step-by-step solution

Given Information and Formula used

Given :

The diameter of drops of ink : $30\mu m$

Charge on the droplets : $800,000$

Speed : 20 m/s.

Dimensions of electrodes : $6.0mm\mathrm{long},4.0mm$wide, $1.0mm$apart.

The distance from the center of the electrodes to the paper : $2.0cm$.

Height of the tallest letters :$6.0mm$

The drops are deflected upward (or downward) by :$3.0mm$

Ink has a density of : $800kg/m^3$

Theory used :

Mass : $m=\rho V$

Maximum deflection : $\tan \theta =\frac{v_y}{v_x}$

Newton's law : $v_{1y}=v_{0y}+a_y(t_1-t_0)$

Field strength :$E=\frac{m{a}_y}{q}$

Calculating the maximum deflection

First, we must locate mass :

$$\begin{gathered} m=\rho V \\ =\rho \left(\frac{4\pi }{3}{r}^3\right) \\ =(800kg/m³)(\frac{4\pi }{3}{(30x{10}^{-6}m)}^3) \\ =\underline{9.04x{10}^{-11}kg} \end{gathered}$$

Then there's charge $q$:

$$\begin{gathered} q=(8x{10}^5)(1.6\times {10}^{-19})C \\ =\underline{1.28x{10}^{-13}C} \end{gathered}$$

We now have a value for maximum deflection $v_{1y}$

role="math" localid="1649140142801" $$\begin{gathered} \tan \theta =\frac{v_y}{v_x} \\ =\left(\frac{2mm}{30mm}\right) \\ =0.15 \\ \Rightarrow v_{1y}=\tan \theta \left(V_x\right) \\ =0.15(20m/s) \\ =\underline{3.0m/s} \end{gathered}$$

Calculating the required electric field strength needed between the electrodes 

We can now calculate acceleration ay using the following relation:

$$\begin{gathered} v_{1y}=v_{0y}+a_y(t_1-t_0) \\ \Rightarrow a_y=\frac{v_{1y}-v_{0y}}{(t_1-t_0)} \\ =\frac{3.0m/s-0m/s}{{\displaystyle \frac{6.0\times {10}^{-3}}{20m/s}}} \\ =\frac{3.0m/s}{3.0\times {10}^{-4}s} \\ \Rightarrow a_y=\underline{1.0\times {10}^{4}m/s^2} \end{gathered}$$

Finally, field strength $E$:

$$\begin{gathered} E=\frac{m{a}_y}{q} \\ =\frac{(9.04x{10}^{-11}kg)(1.0x{10}^{4}m/s^2)}{1.28x{10}^{-13}C} \\ =\boxed{\mathbf{7}\mathbf{.}\mathbf{06}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{N}\mathbf{/}\mathit{C}\mathbf{}} \end{gathered}$$