# 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?

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!

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

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