In a region where there is an electric field, the electric forces do $+8.0 \times 10^{-19} \mathrm{J}\( of work on an electron as it moves from point \)X$ to point \(Y\). (a) Which point, \(X\) or \(Y\), is at a higher potential? (b) What is the potential difference, \(V_{Y}-V_{X},\) between point \(Y\) and point \(X ?\)

Since positive work is done on the electron moving from point X to point Y, it means that the electron is moving in the opposite direction of the electric field (electrons have a negative charge). Therefore, the electron is moving from the lower potential (point X) to the higher potential (point Y). Answer (a): Point Y is at a higher potential.

We know that the work done on the electron is equal to the change in potential energy, which can be related to the potential difference: \begin{align*} W &= q(V_{Y} - V_{X})\\ \end{align*} Where \(W\) is the work done, \(q\) is the charge of the electron, and \(V_Y\) and \(V_X\) are the potentials at points Y and X, respectively. We are given that the work done is \(8.0 \times 10^{-19} \mathrm{J}\), and the charge of an electron is \(-1.6 \times 10^{-19} \mathrm{C}\). We can now calculate the potential difference: \begin{align*} V_{Y} - V_{X} &= \frac{W}{q}\\ &= \frac{8.0 \times 10^{-19}\mathrm{J}}{-1.6 \times 10^{-19}\mathrm{C}}\\ &= -5 \mathrm{V} \end{align*} Answer (b): The potential difference between point Y and point X is \(-5 \mathrm{V}\).