Question

We want to get a better, more concrete idea of the strength of the electric force.   So imagine you could remove all the electrons from 1 mL (1 cubic centimeter) of water.   Dump the electrons on yourself, and put the electron-free nuclei (protons and neutrons) on the moon.   Estimate how much force will there be between the positive and negative charges?  (Of course you have to look up the earth-moon distance.)   What fraction of your weight is that force?

(How many electrons do you have in 1 cc? Consider that virtually all the mass is given by protons and neutrons. Since the proton and the neutron have almost equal masses, and most atoms have equal numbers of protons and neutrons, so the water mass, divided by 2, is the mass of protons. Once you know the total mass of protons, divide by the mass of a single proton and you have the total number of protons. Knowing that, how many electrons do you then suppose you have?)

Answer 1

mass of the water in \(1 \mathrm{~L}=1 \mathrm{~kg}\)

mass of water in \(1 \mathrm{~mL}=1 \mathrm{~kg} \times 10^{-3}=1 \mathrm{gm}\)

from the hint, the mass of protons in \(1 \mathrm{~mL}\) water \(=\frac{1 \mathrm{gm}}{2}=0.5 \mathrm{gm}\) (assume mass of neutrons = mass of protons)

number of protons which make \(=0.5 \mathrm{gm}\) \(\mathrm{N}_{\mathrm{p}}=\frac{0.5}{\mathrm{~m}_{\mathrm{p}}}=2.9893 \mathrm{e}+026\)

hence equal number of electrons should present in \(1 \mathrm{ml}\) of water \(\mathrm{N}_{\mathrm{e}}=2.9893 \mathrm{e}+026\)

the electric force Np number of protons and Ne number of electrons

let distance of seperation \(\mathrm{R}=384,400 \mathrm{~km}\)

$$ \mathrm{F}=\frac{\mathrm{k}\left(\mathrm{q} \mathrm{N}_{\mathrm{p}}\right)\left(\mathrm{q} \mathrm{N}_{\mathrm{e}}\right)}{\mathrm{R}^{2}} ; \mathrm{k} \text { is coloumb constant }=8.98 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} \cdot \mathrm{C}^{-2} $$

substituting all the know values gives \(\mathrm{F}=1.3952 \mathrm{e}+008 \mathrm{~N}\)

weight of average human body \(\mathrm{F}_{\text {men }}=\mathrm{mg}=62 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s} 2=607.6 \mathrm{~N}\)

the ratio \(=\frac{\mathrm{F}}{\mathrm{F}_{\mathrm{men}}}=\frac{1.3952 \mathrm{e}+008 \mathrm{~N}}{607.6 \mathrm{~N}}=2.2962 \mathrm{e}+005\) times

surprisingly the electric force is \(2.2962 \mathrm{e}+005\) times the gravity force on average human