$A$ charge $Q$ is to be divided between two objects. The values of the charges on the objects so that the electrostatic force between them will be maximum is

$ABC$ is a right-angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi / 2$. The three charges $+15\,e.s.u.$,$+12\,e.s.u.$,and $-20\,e.s.u.$ are placed respectively on $A$,$B$,and $C$. The force acting on $B$ is.......$dynes$.

The repulsive force between two particles of same mass and charge,separated by a certain distance,is equal to the weight of one of them. The distance between them is . . . . . . $\times 10^{-1} \ m$.
Mass of particle $= 1.66 \times 10^{-27} \ kg$
Charge of particle $= 1.6 \times 10^{-19} \ C$
$k = 9 \times 10^9 \ MKS, \ g = 10 \ ms^{-2}$

Point charges $+4q, -q$ and $+4q$ are kept on the $x$-axis at points $x = 0, x = a$ and $x = 2a$ respectively. Then:

The sum of two point positive charges separated by a distance of $1.5 \ m$ in air is $25 \mu C$. If the electrostatic force between the two charges is $0.6 \ N$,then the difference between the two charges is (in $\mu C$)

$F$ is the force between two identical charged particles placed at a distance $Y$ from each other. If the distance between the charges is reduced to half the previous distance,then the force between them becomes: