1. **Problem Statement:** Calculate the electric potential $V$ due to a point charge $q_1 = 25.0 \times 10^{-9}$ C at distances (a) 1.00 m, (b) 2.00 m, and (c) infinitely far away. 2. **Formula:** The electric potential $V$ at a distance $r$ from a point charge $q$ is given by: $$V = \frac{kq}{r}$$ where $k = 8.99 \times 10^9$ N\cdot m$^2$/C$^2$ is Coulomb's constant. 3. **Important notes:** - The potential decreases with distance. - At infinite distance, the potential is zero. 4. **Calculations:** (a) At $r = 1.00$ m: $$V = \frac{8.99 \times 10^9 \times 25.0 \times 10^{-9}}{1.00} = 8.99 \times 10^9 \times 25.0 \times 10^{-9}$$ Simplify powers of ten: $$= 8.99 \times 25.0 \times 10^{9 - 9} = 8.99 \times 25.0 = 224.75 \text{ volts}$$ (b) At $r = 2.00$ m: $$V = \frac{8.99 \times 10^9 \times 25.0 \times 10^{-9}}{2.00}$$ Write as: $$V = \frac{8.99 \times 25.0}{\cancel{2.00}} \times 10^{9 - 9} \times \frac{1}{\cancel{2.00}}$$ Canceling 2.00 in denominator: $$V = \frac{224.75}{2} = 112.375 \text{ volts}$$ (c) At infinite distance $r \to \infty$: $$V = \frac{kq}{\infty} = 0 \text{ volts}$$ 5. **Final answers:** - (a) $V = 224.75$ volts - (b) $V = 112.375$ volts - (c) $V = 0$ volts This shows how electric potential decreases with distance from the charge.