Two identical conducting spheres $\mathrm{P}$ and $\mathrm{S}$ with charge $Q$ on each, repel each other with a force $16 \mathrm{~N}$. A third identical uncharged conducting sphere $\mathrm{R}$ is successively brought in contact with the two spheres. The new force of repulsion between $\mathrm{P}$ and $\mathrm{S}$ is :

Three point charges are placed at the corners of an equilateral triangle. Assuming only electrostatic forces are acting

Two point charges $A$ and $B$, having charges $+Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $\mathrm{F}$. If $25 \%$ charge of $A$ is transferred to $B$, then force between the charges becomes

The radius of two metallic spheres $A$ and $B$ are ${r_1}$ and ${r_2}$ respectively $({r_1} > {r_2})$. They are connected by a thin wire and the system is given a certain charge. The charge will be greater

The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $=6.7 \times 10^{-11} \,Nm ^2 / kg ^2$, mass of an electron $=9.1 \times 10^{-31} \,kg$, charge on an electron $=1.6 \times 10^{-19} C$ )