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Asked in GUJCET 2008 · Coulomb's law and superposition
Idea: at a fixed distance, F∝ q₁q₂.
New charges: q₁'=1.1 q₁ and q₂'=0.9 q₂.
(F')/F=((1.1 q₁)(0.9 q₂))/(q₁q₂)=0.99
F'=0.99×200=198 N
The two percentage changes multiply rather than add, so the force falls by 1% instead of staying the same.
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