Two charges $2 C$ and $6 C$ are separated by a finite distance. If a charge of $-4 C$ is added to each of them, the initial force of $12 \times 10^3 \,N$ will change to

Two point charges $+3\,\mu C$ and $+8\,\mu C$ repel each other with a force of $40\,N$. If a charge of $-5\,\mu C$ is added to each of them,then the force between them will become....$N$

In a hydrogen-like system,the ratio of the Coulomb force to the gravitational force between an electron and a proton is of the order of:

Two charges $+10 \mu C$ and $-10 \mu C$ are separated by $10 \text{ cm}$. The magnitude of the force acting on another charge $5 \mu C$ placed at the midpoint of the line joining the two charges will be: [Use $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$] (in $\text{ N}$)

If the Coulombian force acting between two protons separated by a distance $r$ is $F$, what would be the force acting between two alpha particles separated by a distance $2r$?

Two point charges $+8 q$ and $-2 q$ are located at $x=0$ and $x=L$ respectively. The location of a point on the $x$-axis from the origin,at which the net electric field due to these two point charges is zero is