Q.

Two point charges repel each other with a force of 100N. One of the charges is increased by 20% another is reduced by 20%. The new force of repulsion at the same distance would be 

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a

144N

b

96N

c

100N

d

199N

answer is C.

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Detailed Solution

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F=k.Q.Qr2
F'=k×0.8q×1.2qr2
FF'=10.96
F'=96N

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