1. **State the problem:** Calculate the electric field $E$ at a distance $r = 5$ m from a particle with charge $q = 2$ nC (nanocoulombs). 2. **Formula used:** The electric field due to a point charge is given by Coulomb's law: $$E = \frac{k |q|}{r^2}$$ where $k = 8.99 \times 10^9$ N·m²/C² is Coulomb's constant. 3. **Convert units:** Convert charge from nanocoulombs to coulombs: $$2\ \text{nC} = 2 \times 10^{-9}\ \text{C}$$ 4. **Substitute values:** $$E = \frac{8.99 \times 10^9 \times 2 \times 10^{-9}}{5^2}$$ 5. **Simplify denominator:** $$5^2 = 25$$ 6. **Calculate numerator:** $$8.99 \times 10^9 \times 2 \times 10^{-9} = 8.99 \times 2 = 17.98$$ 7. **Calculate electric field:** $$E = \frac{17.98}{25}$$ 8. **Simplify fraction:** $$E = 0.7192\ \text{N/C}$$ **Final answer:** The electric field at 5 m from the particle is approximately $0.72$ N/C.