Question 21.68: Two engineering students, John with a mass of 90 kg and Mary......
Two engineering students, John with a mass of 90 kg and Mary with a mass of 45 kg, are 30 m apart. Suppose each has a 0.01 % imbalance in the amount of positive and negative charge, one student being positive and the other negative. Find the order of magnitude of the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.
The net charge carried by John whose mass is m is roughly
\begin{aligned} q & =(0.0001) \frac{m N_A Z e}{M} \\ & =(0.0001) \frac{(90\, kg )\left(6.02 \times 10^{23} \text { molecules } / mol \right)(18 \text { electron proton pairs } / \text { molecule })\left(1.6 \times 10^{-19} \,C \right)}{0.018 \,kg / mol } \\ & =8.7 \times 10^5\, C , \end{aligned}
and the net charge carried by Mary is half of that. So the electrostatic force between them is estimated to be
F \approx k \frac{q(q / 2)}{d^2}=\left(8.99 \times 10^9\, N \cdot m ^2 / C ^2\right) \frac{\left(8.7 \times 10^5\, C \right)^2}{2(30\, m )^2} \approx 4 \times 10^{18} \,N .