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The force between two charges 0.06 m apart is 5 N. If each charge is moved towards the other by 0.01 m, then the force between them will become

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22.50 N

7.20 N

11.25 N

45.00 N

answer is B.

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Detailed Solution

$F \propto \frac{1}{r^{2}} \Rightarrow \frac{F_{1}}{F_{2}} = {(\frac{r_{2}}{r_{1}})}^{2} \Rightarrow \frac{5}{F_{2}} = {(\frac{0 .04}{0 .06})}^{2} = F_{2} = 11 .25 N$

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