\begin{problem*}{20.3}
A uniform electric field of magnitude $E = 250\U{V/m}$ is directed in
the positive $x$ direction (\ihat).  A $q = +12.0\U{$\mu$C}$ charge
moves from the origin to the point $(x,y) = (20.0\U{cm}, 50.0\U{cm})$.
\Part{a} What is the change in the potential energy $\Delta U$ of the
charge-field system?
\Part{b} Through what potential difference $\Delta V$ does the charge move?
\end{problem*} % problem 20.3

\begin{solution}
\Part{a}
From the text Equation 20.1 (page 643) we see
\begin{equation}
 \Delta U = -q \int_A^B \vect{E} \cdot d\vect{s}
          = -q \int_A^B E\ihat \cdot d\vect{s}
          = -q E \int_{x_1}^{x_2} dx
	  = -q E \Delta x
\end{equation}
Which is the same process the book used to get to their Equation 20.9.
Plugging in our numbers
\begin{equation}
 \Delta U = -12.0\E{-6}\U{C} \cdot 250\U{V/m} \cdot 0.200\U{m}
          = \ans{-6.00\E{-4}\U{J}}
\end{equation}

\Part{b}
The change in electric potential is given by
\begin{equation}
 \Delta V = \frac{\Delta U}{q} = \ans{-50.0\U{V}}
\end{equation}
\end{solution}
