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Reviewed CSCA Physics question · Standard

在真空中，两个点电荷分别带 +2×10⁻⁶ C 和 -4×10⁻⁶ C，相距0.3 m。求它们之间的静电力大小（静电力常量 k=9×10⁹ N·m²/C²）。

  1. 0.2 N
  2. 0.4 N
  3. 0.8 N
  4. 1.6 N
Show the answer and explanation

Correct answer

C. 0.8 N

Principle or equation

库仑定律：F = k|q1q2|/r²。

Why this answer is correct

F = 9×10⁹ × |(2×10⁻⁶)(-4×10⁻⁶)| / (0.3)² = 9×10⁹ × 8×10⁻¹² / 0.09 = 72×10⁻³ / 0.09 = 0.8 N。

Example

若电荷量各为1×10⁻⁶ C，距离0.1 m，则 F = 9×10⁹ × 10⁻¹² / 0.01 = 0.9 N。

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