Two equal point charges q each are placed at distance d. The electrostatic PE of the system (zero at infinity) is:
Ak q² / d
Bk q² / d²
Ck q / d
Dk q²·d
Answer & Solution
Correct answer: A. k q² / d
1. Work needed to bring 2nd charge from ∞ to distance d in field of 1st = k q · q / d.
2. So U = k q₁ q₂ / d = k q² / d (with both = q).
3. For N point charges, sum over all pairs: U = Σᵢⱼ k qᵢqⱼ / rᵢⱼ.
_Source: syllabus Physics Ch 1 §1.13 Potential Energy of a System_
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