Charges of +3.0 μC and −4.0 μC are 0.30 m apart. Taking k = 9 × 109 N m2 C−2, find the force between them in a vacuum, and in an oil of relative permittivity 2.0.
- q1q2 = 3.0 × 10−6 × 4.0 × 10−6 = 1.2 × 10−11 C2, and r2 = 0.302 = 0.090 m2.
- F = 9 × 109 × 1.2 × 10−11 ÷ 0.090 = 0.108 ÷ 0.090 = 1.2 N.
- In the oil: F = 1.2 ÷ 2.0 = 0.60 N.
Answer: 1.2 N in a vacuum and 0.60 N in the oil, attractive in both cases