Question

Two point charges Q1 and Q2 are 2.10 m apart, and their total charge is 21.0 μC. If the force of repulsion between them is 0.206 N, what are magnitudes of the two charges? Enter the smaller charge in the first box.

Answer 1

Given total charge = 21 micro c Q1 and Q 2 are 2.10 m apart force of repulsion = 0.206 N The electrostatic force between the two charges is F = k Q 1 Q 2/R^2 0.0206 N = (9 times 10^9 N m^2/c^2) Q 2 (21 times 10^-6 - Q 2)/(2.10m)^2 = > Q 2 (21 times 10^-6 - Q 2) = 0.206 times (2.10)^2/9 times 10^9 = > 21 times 10^-6 Q 2 - Q^2 2 = 10.094 times 10^-11 = > Q^2_2 - 21 times 10^-6 Q 2 + 10.094 times 10^-11 = 0 rightarrow It is a quadratic equation s0, Q 2 = 21 times 10^-6 plusminus squareroot (21 times 10^-6)^2 - 4 (10.094 times 10^-11)/2 for positive sign Q_2 = 21 times 10^-6 + squareroot 441 times 10^-12 - 40.376 times 10^-11/2 = 21 times 10^-6 + 10^-6 squareroot 37.240/2 = 10^-6 (21 + 6.102458)/2 = 13.55 micro c