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Question

Two similar very small conducting spheres having charges 40C and −20C are at some distance apart. Now they are touched and then kept at a same initial distance. The ratio of the initial to the final force between them is:

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Solution

The correct option is **A** 8:1

Initially force between conducting spheres F1=Kq1q2r2=K(40)(−20)r2(i)

After contact total charge=40−20=20μC

Now charge will be equally distributed

so q1=q2=202=10μC

Finally force between spheresF2=K(10)(10)r2(ii)

Ratio of forces=F1F2=40×2010×10=8

Initially force between conducting spheres F1=Kq1q2r2=K(40)(−20)r2(i)

After contact total charge=40−20=20μC

Now charge will be equally distributed

so q1=q2=202=10μC

Finally force between spheresF2=K(10)(10)r2(ii)

Ratio of forces=F1F2=40×2010×10=8

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