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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.
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Solution
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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