If two charges $q _1$ and $q _2$ are separated with distance ' $d$ ' and placed in a medium of dielectric constant $K$. What will be the equivalent distance between charges in air for the same electrostatic force?

Force of attraction between two point charges $Q$ and $-Q$ separated by $d\,$ metre is ${F_e}$. When these charges are placed on two identical spheres of radius $R = 0.3\,d$ whose centres are $d\,$ metre apart, the force of attraction between them is

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 $=1.6 \times 10^{-19} C$ )

Two free positive charges $4q$ and $q$ are a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed form charge $q$ ?

Four identical pendulums are made by attaching a small ball of mass $100 \,g$ on a $20 \,cm$ long thread and suspended from the same point. Now, each ball is given charge $Q$, so that balls move away from each other with each thread making an angle of $45^{\circ}$ from the vertical. The value of $Q$ is close to …………..$\mu C$ $\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right.$ in $SI$ units $)$