Answer
(a) The point Y is at a higher potential.
(b) $V_y-V_x = 5.0~J/C$
Work Step by Step
(a) We can write an expression for the work done by the electric field:
$W = E~d~q$
$W = -\Delta V~q$
Since the electron's charge is negative and the work is positive, the value of $\Delta V$ must be positive. Therefore, $V_y-V_x$ is positive, so the point Y is at a higher potential.
(b) We can find $V_y-V_x$, which is equal to $\Delta V$:
$W = -\Delta V~q$
$\Delta V = -\frac{W}{q}$
$\Delta V = -\frac{8.0\times 10^{-19}~J}{-1.6\times 10^{-19}~C}$
$V_y-V_x = 5.0~J/C$
