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The electrical force between two point charges is 200 N. If one charge is increased by 10% and the other is decreased by 10%, the electrical force between them at the same distance becomes

Asked in GUJCET 2008 · Coulomb's law and superposition

Answer: (3) 198 N

Step-by-step solution

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.

Why the other options are wrong

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