If two charges $q_1$ and $q_2$ are separated by a distance '$d$' and placed in a medium of dielectric constant $K$,what will be the equivalent distance between the charges in air for the same electrostatic force?

Two point charges $+Q$ and $+q$ repel each other with a force of $100 \,N$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $10 \%$, the force of repulsion between them will:

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 point charges $+q_{1}$ and $+q_{2}$ are placed a finite distance '$d$' apart. It is desired to put a third charge $q_{3}$ in between these two charges so that $q_{3}$ is in equilibrium. This is

If the distance between two point charges $+Q$ and $-Q$ is doubled,what will be the magnitude of the electrostatic force between them?

Electric charges of $1\,\mu C$, $-1\,\mu C$, and $2\,\mu C$ are placed in air at the corners $A$, $B$, and $C$ respectively of an equilateral triangle $ABC$ having a side length of $10\,cm$. The resultant force on the charge at $C$ is......$N$