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The Coulombian repulsive force between two alpha particles kept at a distance of 3 cm in air is _____ N.

Asked in GSEB Board March 2023 · Coulomb's law and superposition

• (1) 1.024×10⁻²⁷
• (2) 1.024×10⁻²⁵
• (3) 1.024×10⁻²⁴
• (4) 1.024×10⁻²³

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives 1.024×10⁻²⁷, which is 1000 times too small. Sanity check: (kq²)/(r²) = ((9×10⁹)(1.024×10⁻³⁷))/(9×10⁻⁴) = 1.024×10⁻²⁴ N.
• (2) This comes from squaring 0.03 m as 0.009 instead of 0.0009: a denominator ten times too big gives a force ten times too small.
• (4) No single slip gives 1.024×10⁻²³, which is ten times too large. Sanity check: q² = (3.2×10⁻¹⁹)² = 1.024×10⁻³⁷ C² and k/(r²) = (9×10⁹)/(9×10⁻⁴) = 10¹³, so F = 1.024×10⁻²⁴ N.

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