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प्रश्न
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.
उत्तर
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,
\[= 9 \times {10}^9 \times \frac{5 . 34 \times {10}^6 \times 5 . 34 \times {10}^6}{{10}^{- 2}}\]
\[ = 2 . 56 \times {10}^{25} N\]
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 ).
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