    Question

# Two identical small conducting spheres carry charges of Q1 and Q2 with Q1>>Q2. The spheres are d distance apart. The force they exert on each other is F1. The spheres are made to touch each other and then separated by a distance d. The force they exert on each other now is F2. Then F1/F2 is :-

A
4Q1Q2
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B
Q14Q2
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C
4Q2Q1
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D
Q24Q1
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Solution

## The correct option is C 4Q2Q1F=KQ1Q2d2Q′1+Q′2=Q1+Q2........(1)KQ′1R1pt1=KQ′2R2⇒Q′1R1=Q′2R2⇒Q′1=(Q1+Q2)−Q′1R2R1⇒Q′1(R1+R2R1)=(Q1+Q2)⇒Q′1=(Q1+Q2)R1(R1+R2)=(Q1+Q2)2⇒Q′2=(Q1+Q2)R2(R1+R2)=Q1+Q22F2=KQ11Q22d2=Kd2(Q1+Q2)24≈KQ214d2∴F1F2=KQ1Q2d2×4d2KQ21=4Q2Q1Hence,optionCiscorrectanswer.  Suggest Corrections  0      Similar questions  Related Videos   The Law of Conservation of Mechanical Energy
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