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JEE Main 2024 (31 Jan, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from JEE Main 2024 (31 Jan, Shift 2)

Q28JEE Main 2024 · 31 JAN · SHIFT 2Easy

Electric Charge & Coulomb's Law

Force between two point charges $q_1$ and $q_2$ placed in vacuum at ' $r$ ' cm apart is $F$ . Force between them when placed in a medium having dielectric constant $K = 5$ at ' $r/5$ ' $\text{cm}$ apart will be:

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Answer: (a)

$F' = \dfrac{1}{K}\dfrac{kq_1q_2}{(r/5)^2} = \dfrac{25}{K}F$
With $K = 5$ : $F' = 5F$
So the answer is (a) $5F$ .

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