Design of an Inkjet Printer. Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.4x10-8 g. The drops will leave the nozzle and travel toward the paper at 50 m/s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude \({\bf{8}}.{\bf{0}} \times {10^4}\) N/C. Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 m/s, what q value is needed to achieve the same 0.30-mm deflection?

a)

The electric force between the two plates responsible for the motion of drop is:

\(\begin{aligned}qE = ma\\q = \frac{{ma}}{E}\end{aligned}\)

Where, \(q\) is the charge of the ink-drop and \(m\) is the mass of the drop. \(E\) is the electric force between the two plates and acceleration is \(a\).

Now, it is given that the drop deflects a distance of \(x = 0.02m\)and with a speed of \(v = 50.0m/s\).

Therefore, the time taken to travel the distance is

\(\begin{aligned}t &= \frac{x}{v}\\ &= \frac{{0.02}}{{50.0}}\\ &= 4.0 \times {10^{ - 4}}s\end{aligned}\)

b)

Now, from Newtons law of motion we calculate the acceleration

\(\begin{aligned}d &= {v_ \circ }t + \frac{1}{2}a{t^2}\\d &= \frac{1}{2}a{t^2}\\a &= \frac{{2d}}{{{t^2}}}\end{aligned}\)

Here \({v_ \circ }\)\( = \)\(0\)and the deflected distance \(d\) is equals to \(0.30mm\). Now substituting we get:

\(q = \frac{{2md}}{{{t^2}E}}\)

Putting the values, we get:

\(\begin{aligned}q &= \frac{{2md}}{{{t^2}E}}\\ &= \frac{{2(1.4 \times {{10}^{ - 11}}kg)(0.30 \times {{10}^{ - 3}}m)}}{{{{(4.0 \times {{10}^{ - 4}}s)}^2}(8.0 \times {{10}^4}N/C)}}\\ &= 6.56 \times {10^{ - 13}}C\end{aligned}\)

Now, the number of electrons is

\(\begin{aligned}n &= \frac{q}{e}\\ &= \frac{{6.56 \times {{10}^{ - 13}}C}}{{1.6 \times {{10}^{ - 19}}C}}\\ &= 4101 \times {10^3}\end{aligned}\)

Therefore, the total number of electrons is \(4101 \times {10^3}\)electrons

Now, when the speed becomes \(v = 25.0m/s\)

The time taken with the new velocity is given by: \({t_2} = \frac{x}{{{v_2}}} = \frac{{0.02m}}{{25.0}} = 8.0 \times {10^{ - 4}}s\)

Now, putting the values in \(m,d,{t_2},{\rm{ }}E\) we get:

\(\begin{aligned}q &= \frac{{2md}}{{{t^2}E}}\\ &= \frac{{2(1.4 \times {{10}^{ - 11}})(0.30 \times {{10}^{ - 3}})}}{{{{(8.0 \times {{10}^{ - 4}})}^2}(8.0 \times {{10}^4})}}\\ &= 1.64 \times {10^{ - 13}}C\end{aligned}\)

Therefore, the charge needed to be taken is \(1.64 \times {10^{ - 13}}C\)