Question

Suppose all the electrons of 100 g water are lumped together to form a negatively-charged particle and all the nuclei are lumped together to form a positively-charged particle. If these two particles are placed 10.0 cm away from each other, find the force of attraction between them. Compare it with your weight.

Answer 1

Molecular mass of water= 18 g
So, number of atoms in 18 g of H₂O = Avogadro's number
= 6.023 × 10²³
Number of electrons in 1 atom of H₂O = (2 × 1) + 8 = 10
Number of electrons in 6.023 × 10²³ atoms of H₂O= 6.023 × 10²⁴
That is, number of electrons in 18 g of H₂O = 6.023 × 10²⁴
So, number of electrons in 100 g of H₂O = $\frac{6.023\times {10}^{24}}{18}\times 100$
= 3.34 × 10²⁵
Total charge = 3.34 × 10²⁵ × (−1.6 × 10⁻¹⁹)
=− 5.34 × 10⁶ C
So total charge of electrons in 100 gm of water, q₁ = −5.34 × 10⁶ C
Similarly, total charge of protons in 100 gm of water, q₂ = +5.34 × 10⁶ C
Given, r = 10 cm = 0.1 m
By Coulomb's Law, electrostatic force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$
$$\begin{gathered} =9\times {10}^9\times \frac{5.34\times {10}^6\times 5.34\times {10}^6}{{10}^{-2}} \\ =2.56\times {10}^{25}\mathrm{N} \end{gathered}$$

This force will be attractive in nature.
Result shows that the electrostatic force is much stronger than the gravitational force between any us and earth(weight=gravitational force between us and earth).