1. **Problem:** Calculate the change in electric potential energy of a proton moving 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 is given by $$\Delta U = qEd$$ where $q$ is the charge, $E$ is the electric field strength, and $d$ is the displacement along the field. 3. **Important rules:** - The charge of a proton is $q = +1.6 \times 10^{-19}$ C. - Since the proton moves in the direction of the field, $d = 0.50$ m. 4. **Calculation:** $$\Delta U = (1.6 \times 10^{-19}) \times 2000 \times 0.50$$ $$\Delta U = 1.6 \times 10^{-19} \times 1000 = 1.6 \times 10^{-16} \text{ J}$$ 5. **Interpretation:** The electric potential energy of the proton increases by $1.6 \times 10^{-16}$ joules as it moves east along the electric field. Final answer: $$\boxed{1.6 \times 10^{-16} \text{ J}}$$