Question

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 find the force between them

Answer 1

this is the answer
in the attacment

Answer 2

The force between two charges is 7.2 N when each charge is moved towards the other by 0.01 m.

The given data is that the force between two charges is given as 5 N which are at a distance of 0.06 m from each other. So the, if charges come close to each other by 0.01 m, that is the distance between them is decreased, so the force will now be increased as the distance between them is inversely related to the force.  

We have $F=\frac{k q 1 q 2}{r^{2}} \Rightarrow \mathrm{F}=5 \mathrm{N} \Rightarrow 5=\frac{k q 1 q 2}{0.06^{2}}$ initially.  

Then, when the distance is decreased by $0.01 \Rightarrow \mathrm{F}^{\prime}=\frac{k q 1 q 2}{0.05^{2}}$.  

So from first equation we have $k q 1 q 2=5 \times(0.06)^{2}$ which will not change, thereby putting this value into second equation we have $F^{\prime}=\frac{5 \times 0.06 \times 0.06}{0.05 \times 0.05}$

$\Rightarrow F^{\prime}=\frac{36}{5}$

$\bold{\Rightarrow F^{\prime}=7.2 N}$