$A$ charge $Q$ is divided into two parts $q$ and $Q - q$. If the Coulomb repulsion between them when they are separated by a distance $r$ is to be maximum,the ratio of $\frac{Q}{q}$ should be:

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 a distance $r$. $A$ third spherical conductor having the 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

Assume that you have $10^{23}$ carbon atoms. All the nuclei are placed at the North Pole of the Earth and all the electrons are placed at the South Pole of the Earth. (Radius of Earth = $6400 \ km$). The force between the charges is approximately:

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of $37^{\circ}$ with each other. When suspended in a liquid of density $0.7 \text{ g/cm}^3$,the angle remains the same. If the density of the material of the sphere is $1.4 \text{ g/cm}^3$,the dielectric constant of the liquid is . . . . . . $\left(\tan 37^{\circ} = \frac{3}{4}\right)$.

Two particles of equal mass $m$ and equal charge $q$ are separated by a distance of $16 \text{ cm}$. They do not experience any net force. The value of $\frac{q}{m}$ is . . . . . . (where $G$ is the universal gravitational constant and $\epsilon_0$ is the permittivity of free space).

In a right-angled triangle $ABC$,$AB = 3 \ cm$,$BC = 4 \ cm$,and $\angle ABC = \frac{\pi}{2}$. Charges of $+15 \ e.s.u.$,$+12 \ e.s.u.$,and $-20 \ e.s.u.$ are placed at $A$,$B$,and $C$ respectively. What is the net force on the charge at $B$ in $dynes$?