Problem 20.9

20.9.
We know that the change in potential is given as a path integral of the electric field.
$\Delta V = -\int_{A}^{B} \vec{E} \cdot d\vec{S}$
We can split the path from A to B up into the two partial paths from A to C and then C to B.
$\Delta V = -\int_{A}^{C} \vec{E} \cdot d\vec{S} - \int_{C}^{B} \vec{E} \cdot d\vec{S}$
The integral from A to C is in the $\hat{y}$ direction, so taking the dot product of the electric field with $d\hat{y}$ yields
$\int_{A}^{C} \vec{E} \cdot d\vec{S} = (E)(\cos 180{}^{\circ})\int_{-.3}^{.5} dy$
and the integral over the path from C to B involves an integral over $d\hat{x}$. However, the dot product of the electric field with the unit vector in the $\hat{x}$ direction is zero.
$\int_{C}^{B} \vec{E} \cdot d\vec{S} = (E)(\cos 90{}^{\circ})\int_{-.2}^{.4} dx$
So, $\Delta V = - (-E)(.5 - (-.3)) - 0$
$\Delta V = (325)(.8) ~=~ 260 V$