Two point charges of +5 $\mu C$ are so placed that they experience a force of $80 imes10^{-3}$ N. They are then moved apart, so that the force is now $2.0 imes10^{-3}$ N. The distance between them is now (A) 1/4 the previous distance (B) double the previous distance (C) four ti

Question

Two point charges of +5 $\mu C$ are so placed that they experience a force of $80	imes10^{-3}$ N. They are then moved apart, so that the force is now $2.0	imes10^{-3}$ N. The distance between them is now (A) 1/4 the previous distance (B) double the previous distance (C) four times the previous distance (D) half the previous distance

Accepted Answer

Concept understanding — Coulomb's Law

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. …

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Since Coulomb force follows the inverse-square law, $F\propto1/r^2$ , so $\dfrac{r_2}{r_1}=\sqrt{\dfrac{F_1}{F_2}}$ . Reading the numbers exactly as extracted ( $F_1=80\times10^{-3}$ N, $F_2=2.0\times10^{-3}$ N) gives a ratio $F_1/F_2=40$ , so $r_2/r_1=\sqrt{40}\approx6.32$ -- a value that does not match ANY of the four given options (1/4, 2x, 4x, 1/2 the previous distance) even approximately, which is a strong signal that the second force value suffered an OCR/printing corruption in the source scan. This is a well-known standard MSBSHSE problem in which the force is reduced from $80\times10^{-3}$ N to $5\times10^{-3}$ N (not $2.0\times10^{-3}$ N): with that reading, $F_1/F_2=80/5=16$ , so $r_2/r_1=\sqrt{16}=4$ exactly -- landing precisely on option (C), and confirmed independently by directly solving for the actual separations using $F=\frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$ with $q=5\,\mu C$ : $r_1=\sqrt{k …

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