Problem — Fraction of electrons to transfer

Problem

Two people at $1\ \text{m}$. Estimate what fraction of the electrons of one of the two would need to be transferred to the other so that the resulting electric repulsion force equals the weight of a person ($\approx 700\ \text{N}$). Body mass $70\ \text{kg}$, mostly water.

Solution

Step 1 — required charge. Impose that the Coulomb force between the two people (distance $r = 1\ \text{m}$, with equal charge $q$) equals $F = 700\ \text{N}$:

$F = k\,\frac{q^2}{r^2} \;\Longrightarrow\; q = r\sqrt{\frac{F}{k}} = 1\cdot\sqrt{\frac{700}{8{,}99\cdot 10^9}} = \sqrt{7{,}79\cdot 10^{-8}} \approx 2{,}8\cdot 10^{-4}\ \text{C}$

Step 2 — corresponding electrons. Divide by the elementary charge $e = 1{,}6\cdot 10^{-19}\ \text{C}$:

$n = \frac{q}{e} = \frac{2{,}8\cdot 10^{-4}}{1{,}6\cdot 10^{-19}} \approx 1{,}7\cdot 10^{15}\ \text{electrons}$

Step 3 — total electrons in the body. The body ($70\ \text{kg}$) is almost entirely water. Moles of water: $70000\ \text{g}/18\ (\text{g/mol}) \approx 3{,}9\cdot 10^3\ \text{mol}$. Each $\text{H}_2\text{O}$ molecule has $10$ electrons ($8+1+1$):

$N_e \approx 3{,}9\cdot 10^3 \cdot 6{,}02\cdot 10^{23} \cdot 10 \approx 2{,}3\cdot 10^{28}\ \text{electrons}$

Step 4 — fraction:

$\frac{n}{N_e} = \frac{1{,}7\cdot 10^{15}}{2{,}3\cdot 10^{28}} \approx 7\cdot 10^{-14} \sim 10^{-13}$

It suffices to shift about one electron in ten thousand billion: the Coulomb force is enormously stronger than gravity.

$\ev{\sim 10^{-13}}$

Links

Topics: Elettrostatica Concepts: Legge di Coulomb Skills: Stime di Fermi