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# Charges and Electrons: Help Please?

Suppose that 0.63 g of hydrogen is separated into electrons and protons. Suppose also that the protons are placed at the Earth’s North PoleandtheelectronsareplacedattheSouth Pole.

The Coulomb constant is 8.98755 × 10^9N· m^2/C^2

and Avogadro’s number is 6.02214× 10^23mol^-1. What is the resulting compressional force on Earth? The radius of the Earth is 6.37× 10^6m. Answer in units of N.

What a diffucult qs, please help.

### 3 Answers

- Bandagadde SLv 61 decade agoFavorite Answer
No.of atoms in 0.63 g of hydrogen =0.63x6.02214x10^23

=3.794x10^23

This is also equal to no. of protons and electrons.

Force between them is

F= 8.98755x10^9x(3.794x10^23x1.6x10^-19)^2/

(6.37x10^6)^2

= 8.16x10^5 N

- JacyLv 71 decade ago
0.63gH2(1mol/2(1.008g))(6.022x10^23atoms/mol)

= 1.88x10^23 atoms

Therefore the # of protons is equal to 1.88x10^23

The Coulomb electrostatic force is given by:

F = kN²e²/(2r)², where N is the # of protons and

2r is the distance between earths' north/south poles

F = (8.98755x10^9)(1.88x10^23)²(1.6x10^-19)² /

[2(6.37x10^6)]²

=(8.98755)(1.88²)(1.6²)/4(6.37²)10^9(10^46)(10^-38)(10^-12)

F = 0.5x10^5 N

- 1 decade ago
1st u find the no of h atoms in .63g

for that 0.63/(6.02214x10^23)

results gives no of h atom

then you coulomb's law of force.

where r=6.37x10^6