1. **Problem:** Calculate the change in electric potential energy when a proton moves 0.50 m east in a uniform electric field of magnitude 2,000 N/C directed east. 2. **Formula:** The change in electric potential energy $\Delta U$ is given by $$\Delta U = qEd\cos\theta$$ where $q$ is the charge, $E$ is the electric field strength, $d$ is the displacement, and $\theta$ is the angle between the field and displacement directions. 3. **Important rules:** - For a proton, $q = +1.6 \times 10^{-19}$ C. - Since the proton moves east and the field is east, $\theta = 0^\circ$ and $\cos 0^\circ = 1$. 4. **Calculation:** $$\Delta U = (1.6 \times 10^{-19}) \times 2000 \times 0.50 \times 1 = 1.6 \times 10^{-19} \times 1000 = 1.6 \times 10^{-16} \text{ J}$$ 5. **Interpretation:** The positive value means the electric potential energy increases by $1.6 \times 10^{-16}$ joules as the proton moves east along the field. **Final answer:** $$\boxed{\Delta U = 1.6 \times 10^{-16} \text{ joules}}$$