The law governing the force between electric charges is known as

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

Three charges $4q, Q$ and $q$ are in a straight line along the $x$-axis at the positions $x=0 \ m$,$x=\frac{1}{2} \ m$,and $x=1 \ m$ respectively. The resultant force on $q$ will be zero if $Q$ is equal to:

$A$ charge $Q$ is divided into two parts $q$ and $Q-q$ and separated by a distance $R$. The force of repulsion between them will be maximum when:

Two balls with equal charges are in a vessel with ice at $-10^{\circ} C$ at a distance of $25 \ cm$ from each other. On forming water at $0^{\circ} C$,the balls are brought nearer to $5 \ cm$ for the interaction between them to be same. If the dielectric constant of water at $0^{\circ} C$ is $80$,the dielectric constant of ice at $-10^{\circ} C$ is:

$(a)$ Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. What is the mutual force of electrostatic repulsion if the charge on each is $6.5 \times 10^{-7} \; C$? The radii of $A$ and $B$ are negligible compared to the distance of separation.
$(b)$ What is the force of repulsion if each sphere is charged double the above amount,and the distance between them is halved?