A positive charge, q sub 1 , of 5 µC is 3 times 10 to the negative 2 power m west of a positive charge, q sub 2 , of 2 µC What is the magnitude and direction of the electrical force, cap f sub e , applied by q sub 1 on q sub 2 ?
Answer
A
magnitude: 100 cap n direction: east
B
magnitude: 100 cap n direction: west
C
magnitude: 3 cap n direction: west
D
magnitude: 3 cap n direction: east
Explanation
Use Coulomb’s law: F = k|q₁q₂|/r² = (9 × 10⁹)(5 × 10⁻⁶)(2 × 10⁻⁶)/(3 × 10⁻²)² = 100 N. Since both charges are positive, they repel; q₂ is east of q₁, so the force on q₂ points east. The correct choice is magnitude 100 N, direction east.
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