If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them 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 point charge $q_1=4 q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x =12\,cm$. Charge of proton is $q_0$. The proton is placed on $x$-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is $..........cm$.

Two identical conducting spheres carrying different charges attract each other with a force $F$ when placed in air medium at a distance $'d'$ apart. The spheres are brought into contact and then taken to their original positions. Now the two spheres repel each other with a force whose magnitude is equal to that of the the initial attractive force. The ratio between initial charges on the spheres is

Two identical conducting spheres, having charges of opposite sign, attract each other with a force of $0.108$ $N$ when separated by $0.5$ $m$. The spheres are connected by a conducting wire, which is then removed, and thereafter, they repel each other with a force of $ 0.036$ $N$. The initial charges on the spheres are

Two identical conducting spheres with negligible volume have $2.1\, nC$ and $-0.1\, nC$ charges, respectively. They are brought into contact and then separated by a distance of $0.5 \,m$. The electrostatic force acting between the spheres is $.......... \, \times 10^{-9} \,N$

[Given : $4 \pi \varepsilon_{0}=\frac{1}{9 \times 10^{9}} SI$ unit]