The force between two charges $0.06\,m$ apart is $5\,N$. If each charge is moved towards the other by $0.01\,m$,then the force between them will become.........$N$

State Coulomb's law and explain its scalar form.

Two identical balls having like charges $Q_1$ and $Q_2$ are placed at a distance $r$ apart and repel each other with a force $F$. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases $4.5$ times in comparison with the initial value. The ratio of the initial charges of the balls, $\frac{Q_1}{Q_2}$, is:

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:

As shown in the figure,a configuration of two equal point charges $(q_0 = +2 \mu C)$ is placed on an inclined plane. The mass of each point charge is $20 \, g$. Assume that there is no friction between the charge and the plane. For the system of two point charges to be in equilibrium (at rest),the height $h = x \times 10^{-3} \, m$. The value of $x$ is $..........$. (Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, Nm^2 C^{-2}, g = 10 \, ms^{-2}$)

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 $B$?