Two equal point charges exert a force F1 on each other when they are [tex]5.0\times10^{-2}m[/tex] apart in a vacuum. Determine the distance between the charges if the force is to be doubled.

The electric force between two point charges is given as [tex]F\alpha \frac{q1q2}{r^2}[/tex]

where q1 and q2 are the charges, F= force and r = distance apart.

Since the force is inversly proportional to the square of the distance apart, if the force is doubled, then the square of the distance apart will be reduced by 2.

[tex]\rightarrow;r^2=\frac{5.0\times10^{-2}\times5.0\times10^{-2}}{2}= \frac{25\times10^{-4}}{2\times10^0}=1.25\times10^{-3}m[/tex]

But am not too sure of my answer. Please I need help in tackling this problem.