Two point charges $A$ and $B$ with magnitudes $+4q$ and $-4q$ are placed along a line separated by a distance $r$. The force acting between them is $F$. If $25\%$ of the charge from point $A$ is transferred to point $B$,the force between the charges now becomes:

Two charged spheres separated at a distance $d$ exert a force $F$ on each other. If they are immersed in a liquid of dielectric constant $2$,then what is the force (if all conditions are same)?

If the Coulombian force acting between two protons separated by a distance $r$ is $F$, what would be the force acting between two alpha particles separated by a distance $2r$?

Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x$ $(x \ll a)$ along $X$-axis. The force acting on the point charge is ($x$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

Two point charges $A$ and $B$,having charges $+Q$ and $-Q$ respectively,are placed at a certain distance apart and the force acting between them is $F$. If $25 \%$ of the charge of $A$ is transferred to $B$,then the force between the charges becomes

Two point charges of $+2 \, \mu \text{C}$ and $+6 \, \mu \text{C}$ repel each other with a force of $12 \, \text{N}$. If a charge of $-4 \, \mu \text{C}$ is added to each,the force will be: