in a certain organic molecule, the nuclei of two carbon atoms are separated by a distance of 0.15nm, what is the magnitude of the electric force between them?
F = kq₁q₂/r² where k = 9·10⁹ N·m²/C² is Coulomb's constant.
Carbon has atomic number 6 (i.e., there are 6 protons in a carbon nucleus). The electric charge on a single proton is e = 1.602·10⁻¹⁹ C. So the electrostatic repulsion between the two carbon nuclei is F = k·(6e)²/(0.15 [nm])² = (9·10⁹ [N·m²/C²])·(6 · 1.602·10⁻¹⁹ [C])²/(0.15·10⁻⁹ [m])² =36864 *10^ -11 N
=3.6864 *10^ -7 N.
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