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.........$N$

Two point charges $+9e$ and $+e$ are placed at a distance of $16\, cm$ from each other. Where should a third charge $q$ be placed between them so that the system remains in equilibrium?

The ratio of the forces between two small spheres with constant charge $(a)$ in air and $(b)$ in a medium of dielectric constant $K$ is:

The force between two point charges $A$ and $B$ is $F$. If $75\%$ of the charge on $A$ is transferred to $B$,then the new force between $A$ and $B$ is:

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: (in $N$)

In one model of the electron,the electron of mass $m_e$ is thought to be a uniformly charged shell of radius $R$ and total charge $e$,whose electrostatic energy $E$ is equivalent to its mass $m_e$ via Einstein's mass-energy relation $E = m_e c^2$. In this model,$R$ is approximately ($m_e = 9.1 \times 10^{-31} \, kg$,$c = 3 \times 10^8 \, ms^{-1}$,$1 / 4 \pi \varepsilon_0 = 9 \times 10^9 \, Nm^2C^{-2}$,magnitude of the electron charge $e = 1.6 \times 10^{-19} \, C$).