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Two point charges repel each other with force $F$. If the distance between them is doubled while the charges are unchanged, the new force is
A$F/2$
B$2F$
C$4F$
D$F/4$
Answer & Solution
Correct answer: D. $F/4$
1. Coulomb's law: $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r^2}$, so $F \propto 1/r^2$.
2. Replacing $r$ by $2r$ multiplies $r^2$ by 4.
3. Hence the force falls to $F/4$.
_Source: Samacheer Kalvi Class 12 Physics Vol.1, Unit 1 'Electrostatics', §1.1 (Coulomb's law)_
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