Answer to Question #167624 in Electric Circuits for edlen kate

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:

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