A small object has charge Q. Charge q is removed from it and placed on a second small object. The two objects are placed 1 m apart. For the force that each object exerts on the other to be a maximum, q should be :-

$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q q_{0}}{\pi^{2}}$ [কুলম্বের সূত্র]

$\begin{array}{l}\text { এখানে, } q=q \\ \begin{array}{l}q_{0}=Q-q \\ r=1 \mathrm{~m}\end{array} \\ \therefore F=k \frac{q(Q-q)}{1} \\\end{array}$

$\Rightarrow F=k\left(Q q-q^{2}\right)$

For $F_{max}$ $; \frac{d F}{d q}=0$

$\begin{array}{l} \Rightarrow \frac{d}{d q}\left(k\left(Q q-q^{2}\right)\right)=0 \\ \Rightarrow k(Q-2 q)=0 \\ \Rightarrow q=\frac{Q}{2} \end{array}$