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,     

\[F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}\]

\[= 9 \times  {10}^9  \times \frac{5 . 34 \times {10}^6 \times 5 . 34 \times {10}^6}{{10}^{- 2}}\] 

\[ = 2 . 56 \times  {10}^{25}   N\]

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 ).