A 10 charge is divided into two parts and placed at 1cm distance so...

Let the charges be $x\mu C, \left(q - x\right)\mu C$, where $q = 10\mu C$

The force between them is

$F = \dfrac{Kx ( q - x )}{{r}^{2}}$.

For the force to be maximum,

$\dfrac{dF}{dx} = 0 ⇒ \dfrac{dF}{dx} = \dfrac{K ( q - 2x )}{{r}^{2}} = 0 ⇒ x = \dfrac{q}{2} = 5\mu C$