What must the distance be between charges of + 2.35 and - 1.96 for the attractive force between them to be the same as that between charges of + 4.06 and - 2.11 separated by a distance of 2.26 pm?

Force is:F = k * (q1 * q2) / r²Therefore:k * (2.35 * 1.96) / r² = k * (4.06 * 2.11) / (2.26) ² / : k4.606 / r² = 8.5666 / 5.10768.5666 r² = 4.606 * 5.1076r² = 23.5256 : 8.5666r² = 2.7462r = √2.7462Answer:r = 1.66 pm