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MHT CET 2024 (10 May, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from MHT CET 2024 (10 May, Shift 2)

Q30MHT CET 2024 · 10 MAY · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two point charges $(A$ and $B) + 4q$ and $-4q$ are placed along a line separated by a distance I '. Force acting between them is F. If $25\%$ of charge from point $A$ is transferred to that at point B, the force between the charges now becomes

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Answer: (c)

$F = \dfrac{k(4q)(4q)}{l^2} = \dfrac{16kq^2}{l^2}$
25% of $4q$ is $q$ . After transfer: $A = +3q$ , $B = -4q + q = -3q$ .
$F' = \dfrac{9kq^2}{l^2} = \dfrac{9}{16}F$
So the answer is (c) $\dfrac{9}{16}F$ .

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