1. **Problem Statement:** Calculate the electric field magnitude at point P, which is the midpoint between two nanoparticles with opposite charges separated by a distance $r = 2.0$ mm. 2. **Relevant Formula:** The electric field $E$ due to a point charge $q$ at a distance $r$ is given by Coulomb's law: $$E = \frac{k |q|}{r^2}$$ where $k = 8.99 \times 10^9$ N m$^2$/C$^2$ is Coulomb's constant. 3. **Important Notes:** - Since point P is the midpoint between two opposite charges, the fields due to each charge add up in magnitude because they point in the same direction. - The distance from each charge to point P is $\frac{r}{2} = 1.0$ mm = $1.0 \times 10^{-3}$ m. 4. **Given Data:** - Electric field magnitude at point P is $120$ N/C. - Distance between charges $r = 2.0$ mm = $2.0 \times 10^{-3}$ m. 5. **Calculate the magnitude of the charge $q$ on each nanoparticle:** Since the fields add, $$E = 2 \times \frac{k |q|}{(r/2)^2}$$ 6. **Simplify the denominator:** $$(r/2)^2 = \left(\frac{2.0 \times 10^{-3}}{2}\right)^2 = (1.0 \times 10^{-3})^2 = 1.0 \times 10^{-6}$$ 7. **Rewrite the equation:** $$120 = 2 \times \frac{8.99 \times 10^9 \times |q|}{1.0 \times 10^{-6}}$$ 8. **Isolate $|q|$:** $$120 = 2 \times 8.99 \times 10^9 \times |q| \times 10^6$$ $$120 = 1.798 \times 10^{16} \times |q|$$ 9. **Solve for $|q|$:** $$|q| = \frac{120}{1.798 \times 10^{16}} = 6.68 \times 10^{-15}$$ 10. **Final answer:** The magnitude of the charge on each nanoparticle is approximately $$|q| = 6.68 \times 10^{-15} \text{ C}$$ This means each nanoparticle carries a very small charge consistent with nanoscale particles.