Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is $q$ and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as:

A metal cube of side $$5 \mathrm{~cm}$$ is charged with $$6 \mu \mathrm{C}$$. The surface charge density on the cube is

The value of electric potential at a distance of $$9 \mathrm{~cm}$$ from the point charge $$4 \times 10^{-7} \mathrm{C}$$ is [Given $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$$] :

A thin spherical shell is charged by some source. The potential difference between the two points $$C$$ and $$P$$ (in V) shown in the figure is:

(Take $$\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$$ SI units)