A charge Q is divide intointo parts and then they are placed at a fixed. The force between the two charges is always maximum when charges are

 

let parts are sepereted at a distance of d

so repulsion force F= k q(Q-q)/d²

     here Q and d are constant so for maximum force differentiate the force w.r.t q

 dF/dq = k/d² [Q -2q]

 for maximum force dF/dq=0

                             Q-2q=0

                              Q/q =2

Dear student

The above ans is incorrect , Please follow this  
Let q1 = qand q2 = Q-q 
r  be the distance or separation between them 
The force of repusion between them is 
F = k (Q-q)q/r^2 
Differenciating F with respect to q and setting it to zero will give us extreme force.
dF/dq = 0 
so, q/Q = 1/2 
For this value q/Q the force is maximum .
The force will be maximum if the second derivative of F is less than 0 
d^2F /dq^2= -2k/r^2 which is less than 0.
Thus the force of repulsion is maximum when q=Q/2 
so , q1 = Q/2 an q2 = Q/2 
Hope this helps