Question

Two unknown charges attract each other. The first charge has twice greater magnitude than the second charge. If the first acting under charge is 20^-12 N and the distance between them is 120 nm, what is the charge of the second charge? what is the force at 200 nm?

Answer 1

$F = k\frac{q_1 \times q_2}{d^2}$

$k = 8.99 \times 10^9 \;N \;m^2/C^2$ is Coulomb constants

$q_1 = 2q_2 \\ 20^{-12} = 8.99 \times 10^9 \frac{2q_2^2}{(120 \times 10^{-9})^2} \\ 20^{-12} = 8.99 \times 10^9 \frac{2q_2^2}{1.44 \times 10^{-14}} \\ 20^{-12} = 12.48 \times 10^{23} q_2 \\ q_2 = 1.39 \times 10^{-20} \;C \\ q_1 = 2.78 \times 10^{-20} \;C$

At 200 nm:

$F = 8.99 \times 10^9\frac{2.78 \times 10^{-20} \times 1.39 \times 10^{-20}}{(200 \times 10^{-9})^2} \\ = 8.99 \times 10^9 \frac{3.86 \times 10^{-40}}{4.0 \times 10^{-14}} \\ = 8.99 \times 10^9 \times 0.965 \times 10^{-26} \\ = 8.67 \times 10^{-17} \;N$