A positive charge q₁ is equal in magnitude to the total electronic charge contained in 0.9 mg of pure water. A second positive charge q₂ is produced on a 6.35 mg copper sphere by removing 0.1% of all its electrons. The two charged objects are separated by 1 km. Use $N_A=6×10²³$ mol⁻¹, M(H₂O)=18 g mol⁻¹, M(Cu)=63.5 g mol⁻¹, atomic number of Cu=29, e=1.6×10⁻¹⁹ C and k=9×10⁹ N m² C⁻². The electrostatic force between them is approximately:
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Question

A positive charge q₁ is equal in magnitude to the total electronic charge contained in 0.9 mg of pure water. A second positive charge q₂ is produced on a 6.35 mg copper sphere by removing 0.1% of all its electrons. The two charged objects are separated by 1 km. Use $N_A=6×10²³$ mol⁻¹, M(H₂O)=18 g mol⁻¹, M(Cu)=63.5 g mol⁻¹, atomic number of Cu=29, e=1.6×10⁻¹⁹ C and k=9×10⁹ N m² C⁻². The electrostatic force between them is approximately:

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G1 Physics · Accepted Answer

For water: n=(0.9×10⁻³)/18=5×10⁻⁵ mol. Each H₂O molecule has 10 electrons, so N₁=5×10⁻⁵×6×10²³×10=3×10²⁰ and q₁=N₁e=48 C. For copper: n=(6.35×10⁻³)/63.5=1×10⁻⁴ mol. Total electrons N=1×10⁻⁴×6×10²³×29=1.74×10²¹. Removing 0.1%=10⁻³ gives N₂=1.74×10¹⁸ and q₂=0.2784 C. Thus F=kq₁q₂/r²=[9×10⁹×48×0.2784]/(10³)²≈1.20×10⁵ N.