Explanation

(a) Let Q is divided into two parts. If one part is q then other will (Q-q).
Let these parts are kept a (Q-q)
distance r.
The electrostatic force between them
$F = K.\frac{q(Q - q)}{r^{2}}\ldots\ldots\ldots(i)$
(where K=constant –$(1/4\pi\varepsilon_{0})$
On differentiating Eq. (i) w.r.t. q, we get
$\frac{dF}{dq} = \frac{d}{dq}\left\lbrack \frac{Kq}{r^{2}}(Q - q) \right\rbrack$
$or\ \ \ \ \frac{dF}{dq} = \frac{K}{r^{2}}\frac{d}{dq}\left\lbrack Qq - q^{2} \right\rbrack$
$or\ \frac{dF}{dq} = \frac{K}{r^{2}}\ \lbrack Q - 2q\rbrack\ \ \ \ldots\ldots(ii)$
But we know that, when force is maximum then
$\ \ \ \ \ \frac{dF}{dq} = 0$
Then from Eq. (ii), we have
$\frac{K}{r^{2}}\lbrack Q - 2q\rbrack = 0\ or\ Q - 2q = 0$
$Or\ \ Q = 2q\ \ \ or\ q = \frac{Q}{2}$
So, the other part $= Q - \frac{Q}{2} = \frac{Q}{2}$
Hence, the each part have the same charge.
$\frac{Q}{2}$