**Problem:** A point charge Q = 5 µC is placed at the origin. Find the electric field magnitude at a point 0.10 m away from the charge using the potential method. **Solution Steps:** 1. Write the expression for electric potential of a point charge: V(r) = k Q / r, where k = 8.99×10^9 N·m²/C². 2. Plug the numbers: V(0.10 m) = (8.99×10^9)(5×10⁻⁶) / 0.10 = 4.495×10^5 V. 3. The electric field is the negative gradient of the potential. In the radial direction for a point charge, E = -dV/dr. 4. Differentiate V = kQ/r with respect to r: dV/dr = -kQ/r². 5. Apply the minus sign: E = -(-kQ/r²) = kQ/r². 6. Compute: E = (8.99×10^9)(5×10⁻⁶) / (0.10)² = (8.99×10^9)(5×10⁻⁶) / 0.01 = 4.495×10^6 N/C. 7. Direction: The field points radially outward because the charge is positive. **Answer:** The electric field magnitude at 0.10 m is 4.5×10⁶ N/C outward.