$12$ positive charges of magnitude $q$ are placed on a circle of radius $R$ in a manner that they are equally spaced. $A$ charge $Q$ is placed at the centre. If one of the charges $q$ is removed,then the force on $Q$ is

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 $= 1.6 \times 10^{-19} \ C$
$k = 9 \times 10^9 \ MKS, \ g = 10 \ ms^{-2}$

$A$ certain charge $Q$ is divided into two parts $q$ and $(Q-q)$. How should the charges $Q$ and $q$ be divided so that $q$ and $(Q-q)$ placed at a certain distance apart experience maximum electrostatic repulsion?

Four charges are arranged at the corners of a square $ABCD$,as shown in the adjoining figure. The force on the charge kept at the centre $O$ is

$A$ charged metallic sphere $A$ is suspended by a nylon thread. Another charged metallic sphere $B$ held by an insulating handle is brought close to $A$ such that the distance between their centres is $10 \, cm$,as shown in Figure $(a)$. The resulting repulsion of $A$ is noted (for example,by shining a beam of light and measuring the deflection of its shadow on a screen). Spheres $A$ and $B$ are touched by uncharged spheres $C$ and $D$ respectively,as shown in Figure $(b)$. $C$ and $D$ are then removed and $B$ is brought closer to $A$ to a distance of $5.0 \, cm$ between their centres,as shown in Figure $(c)$. What is the expected repulsion of $A$ on the basis of Coulomb's law? Spheres $A$ and $C$ and spheres $B$ and $D$ have identical sizes. Ignore the sizes of $A$ and $B$ in comparison to the separation between their centres.

Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them, the initial force of $12 \times 10^3 \,N$ will change to