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Q 1.12(b): What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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We know, Force between two charged particles separated by a distance r is:

F = k\frac{q_{1}q_{2}}{r^2} = k\frac{q^2}{r^2}         (\because q_{1} = q_{2} = q)

Now if q\rightarrow 2q\ and\ r\rightarrow r/2

New value of force:

F_{new} =k\frac{(2q)^2}{(r/2)^2} = 16k\frac{q^2}{r^2} = 16F

Therefore, the force increases 16 times!

F_{new} =16F = 16\times1.5\times10^{-2} N = 0.24\ N

Posted by

HARSH KANKARIA

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