\begin{problem*}{86}
In an inkjet printer, letters are built up by squirting drops of ink
at the paper from a rapidly moving nozzle.  The ink drops, which have
a mass of $1.4\E{-8}\U{g}$ each, leave the nozzle and travel toward
the paper at $20\U{m/s}$, passing through a charging unit that gives
each drop a positive charge $q$ by removing some electrons from it.
The drops then pass between parallel deflecting plates $2.0\U{cm}$
long where there is a uniform vertical electric field with magnitude
$8.0\E{4}\U{N/C}$.  If a drop is to be delected $0.30\U{mm}$ by the
time it reaches the end of the deflection plates, what magnitude of
charge must be given to the drop?
\end{problem*}

\begin{solution}
\begin{center}
\begin{asy}

\end{asy}
\end{center}
From the forces on the drop in each direction
\begin{align}
  F_x &= 0 = m a_x  &
  F_y &= qE = m a_y \\
  a_x &= 0  &
  a_y &= \frac{qE}{m} \;.
\end{align}
Now this looks like a projectile motion problem from your
intro-mechanics class.  No acceleration in the $x$ direction means
$v_x$ is constant, so the time-of-flight is given by.
\begin{align}
  \Delta x &= v_x \Delta t  &
  \Delta t &= \frac{\Delta x}{v_x} = 1\U{ms} \;.
\end{align}
We can plug this time-of-flight into our constant-acceleration
equation for $y(t)$,
\begin{align}
  y(t) &= \frac{a_y}{2}t^2 + v_{y0} t + y_0 \\
  \Delta y &= \frac{qE}{2m} \Delta t^2 \\
  q &= \frac{2m \Delta y}{E \Delta t^2}
    = \frac{2\cdot1.4\E{-11}\U{kg}\cdot3.0\E{-4}\U{m}}{8.0\E{4}\U{N/C}\cdot(1\E{-3}\U{s})^2}
    = \ans{1.05\E{-13}\U{C}}
\end{align}
\end{solution}
