1. **State the problem:** We need to find the distance from a 2 mC charge where the electric field strength is 4 N/C. 2. **Formula used:** The electric field $E$ due to a point charge $Q$ at a distance $r$ is given by: $$E = \frac{k |Q|}{r^2}$$ where $k = 9 \times 10^9$ N m$^2$/C$^2$ is Coulomb's constant. 3. **Given values:** - Charge $Q = 2$ mC $= 2 \times 10^{-3}$ C - Electric field $E = 4$ N/C 4. **Rearrange the formula to solve for $r$:** $$r^2 = \frac{k |Q|}{E}$$ 5. **Substitute the values:** $$r^2 = \frac{9 \times 10^9 \times 2 \times 10^{-3}}{4}$$ 6. **Calculate numerator:** $$9 \times 10^9 \times 2 \times 10^{-3} = 18 \times 10^6$$ 7. **Calculate $r^2$:** $$r^2 = \frac{18 \times 10^6}{4} = 4.5 \times 10^6$$ 8. **Find $r$ by taking the square root:** $$r = \sqrt{4.5 \times 10^6} = \sqrt{4.5} \times \sqrt{10^6} = 2.121 \times 10^3 = 2121 \text{ meters}$$ **Final answer:** The point is approximately 2121 meters from the 2 mC charge where the electric field is 4 N/C.