College Physics (4th Edition)

(a) The point Y is at a higher potential. (b) $V_y-V_x = 5.0~J/C$
(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 $\Delta V$: $W = -\Delta V~q$ $\Delta V = -\frac{W}{q}$ $\Delta V = -\frac{8.0\times 10^{-19}~J}{-1.6\times 10^{-19}~C}$ $V_y-V_x = 5.0~J/C$