NEET-XII-Physics
29: Electric Field and Potential
- #10Suppose 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.Ans : 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}}``
`` =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\,}``
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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