Q.Two charges of 1.0 C each are placed one metre apart in free space. The force between them will be (A) 1.0 N (B) $9\times10^9$ N (C) $9\times10^{-9}$ N (D) 10 N

Coulomb's Law is the single quantitative statement of how two point charges push or pull on each other. Two charges q₁ and q₂ separated by a distance r exert on each other a force of magnitude

F = k · q₁q₂ / r², where k = 1/(4πε₀) ≈ 9 × 10⁹ N·m²·C⁻².

Three ideas live inside that one line, and almost every JEE Main question is really testing one of them.

1 — It is an inverse-square law. The force falls off as 1/r², not 1/r. Double the separation and the force drops to a quarter; halve it and the force quadruples. This is why a graph of F against r is a hyperbola, while F against 1/r² is a straight line through the origin whose slope is k q₁q₂.

2 — It is a vector, and it obeys superposition. The force between two charges points along the line joining them: repulsive for like signs, attractive for unlike signs. When several charges act on one, the net force is the vector sum of the individual Coulomb forces — each computed as if the others were absent. This is the master key to every triangle, square, and collinear-charge problem: never add magnitudes blindly; resolve into components or use the resultant law R = √(F₁² + F₂² + 2F₁F₂cosθ).

3 — Charge is quantized and conserved. Any charge is an integer multiple of the electronic charge, q = ne with e = 1.6 × 10⁻¹⁹ C. When two identical conductors touch, their charge redistributes so each carries the algebraic mean (q₁ + q₂)/2 — signs included. Forgetting the sign here (averaging magnitudes instead) is the most common single mistake in the whole chapter.

The role of the medium. In a medium of dielectric constant K, the force is reduced: F_medium = F_vacuum / K. Equivalently, a separation r in a medium behaves like a larger separation r√K in vacuum. …