Two point charges placed at a certain distance r in air exert force F on each other. Then the distance r₁ at which these charges will exert the same force in a medium of dielectric constant K is (a) r (b) r/K (c) r/√(K) (d) r√(K)

Answer: (c). In air F = (kq₁q₂)/(r²); in medium F = (kq₁q₂)/(K r₁²). Equal forces: K r₁² = r² ⇒ r₁ = r/√(K) So the answer is (c) r/√(K).

Two point charges placed at a certain distance r in air exert force F… | MHT CET 2026 (15 May, Shift 2)

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Electric Charge & Coulomb's Law

Two point charges placed at a certain distance $r$ in air exert force $F$ on each other. Then the distance $r_1$ at which these charges will exert the same force in a medium of dielectric constant $K$ is

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Answer: (c)

In air $F = \dfrac{kq_1q_2}{r^2}$ ; in medium $F = \dfrac{kq_1q_2}{K r_1^2}$ .
Equal forces: $K r_1^2 = r^2 \Rightarrow r_1 = \dfrac{r}{\sqrt{K}}$
So the answer is (c) $\dfrac{r}{\sqrt{K}}$ .

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