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Two point charges repel each other with a force of 100 N. One of the charges isincreased by 10% and other is decreased by 10%. The new force of repulsion at thesame distance will be55 N| 77 N99 N100 N |
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Answer» GIVEN:FORCE=F=100NLet q be the magnitude of two charges.According to Coulomb's law :For constant DISTANCE,Force is directly proportional to product of chargesF1= qxqIf one charge is increased by 10% and the other is reduced by 10% Then New charges will be 0.9q and 1.1 qSo New force will be F2F2=0.9q x 1.1qTake the ratio of F1/F2F1/F2=qxq/0.9qx1.1q=1/0.99=1.01F2=F1/1.01=100/1.01=99NTherefore, the force of repulsion at the same distance is 99NSo option B - 99N is correct answer. the answer is correct b -99N |
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