electrostatics Questions

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges $q$ repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is:

A charged particle $q$ is shot towards another charged particle $Q$ which is fixed. With a speed $v$, it approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$, the closest distance of approach would be:

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium, the value of $q$ is:

A charged oil drop is suspended in a uniform field of $3\times 10^4\text{V/m}$ so that it neither falls nor rises. The charge on the drop will be: (Take mass of the drop $=9.9\times 10^{-15}\text{kg}$ and $g=10{\text{m/s}}^2$)