Three charges $4q$,$Q$,and $q$ are placed in a straight line at positions $0$,$l/2$,and $l$ respectively. The resultant force on $q$ will be zero if $Q = $

The Coulombian repulsive force between two alpha particles kept at a distance of $3 \ cm$ in air is . . . . . . .

Two insulated charged copper spheres $A$ and $B$ have their centres separated by a distance of $50 \; cm$. The charge on each is $6.5 \times 10^{-7} \; C$. Suppose the spheres $A$ and $B$ have identical sizes. $A$ third sphere of the same size but uncharged is brought in contact with the first,then brought in contact with the second,and finally removed from both. What is the new force of repulsion between $A$ and $B$?

If the distance between two equal point charges is doubled,what would happen to the force between them?

When air is replaced by a dielectric medium of constant $k$,the maximum force of attraction between two charges separated by a distance $r$:

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).