Question

Two point charges are separated by a distance r. If the separation is reduced by a factor of 1.5, by what factor does the electric force between them change? 

Answer 1

Now We know that

Electric force

$F=\frac{kq_1q_2}{r^2}$

$F\alpha\frac{1}{r^2}$

Another case

$F'\alpha\frac{1}{r'^2}$

Both equation We can written as

$\frac{F}{F'}=\frac{r'^2}{r^2}$

$r'=\frac{r}{1.5}$

$\frac{F}{F'}=\frac{1}{1.5^2}$

$F'=2.25F$