Two point charges $+8 \mu C$ and $+12 \mu C$ repel each other with a force of $48 \ N$. When an additional charge of $-10 \mu C$ is given to each of these charges (the distance between the charges is unaltered),then the new force is:

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant $K = \frac{1}{4 \pi \varepsilon_0}$ to the gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $= 1.6 \times 10^{-19} \; C$,the mass of the electron $= 9.11 \times 10^{-31} \; kg$,the mass of the proton $= 1.67 \times 10^{-27} \; kg$):

When the distance between the charged particles is halved,the force between them becomes

Four electric charges $2 \mu C, Q, 4 \mu C$ and $12 \mu C$ are placed on the $x$-axis at distances $x=0, 1 \ cm, 2 \ cm$ and $4 \ cm$ respectively. If the net force acting on the charge at the origin is zero,then $Q=$ (in $\mu C$)

Three charges, each of $+1 \mu C$, are placed at the corners of an equilateral triangle. If the repulsive force between any two charges is $F$, then the net force on any one charge will be:

Two point charges $q_1$ and $q_2$ are placed at a distance of $50 \ cm$ from each other in air and interact with a certain force. Now,these charges are placed in an oil with a relative permittivity of $5$. If the force between them remains the same,the distance between them in the oil is ........ $cm$.