在真空中有两个点电荷，电荷量分别为 +2×10⁻⁶ C 和 -3×10⁻⁶ C，相距0.3 m，静电力常量 k = 9×10⁹ N·m²/C²。求它们之间的静电力大小。
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Correct answer
A. 0.6 N
Principle or equation
库仑定律：F = k |q1 q2| / r²。
Why this answer is correct
F = 9×10⁹ * (2×10⁻⁶ * 3×10⁻⁶) / (0.3)² = 9×10⁹ * 6×10⁻¹² / 0.09 = 54×10⁻³ / 0.09 = 0.6 N。
Example
若距离为0.1 m，则 F = 9×10⁹ * 6×10⁻¹² / 0.01 = 5.4 N。
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