$A$ total charge $Q$ is broken into two parts $Q_1$ and $Q_2$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when:

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

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?

If two particles $A$ and $B$ of charges $1.6 \times 10^{-19} \ C$ and $3.2 \times 10^{-19} \ C$ respectively are separated by a distance of $3 \ cm$ in air,then the magnitude of electrostatic force on particle $A$ due to particle $B$ is

Two positive ions,each carrying a charge $q,$ are separated by a distance $d.$ If $F$ is the force of repulsion between the ions,the number of electrons missing from each ion will be ($e$ being the charge on an electron).

Two small spheres,each having an equal positive charge $Q$ (Coulomb),are suspended by two insulating strings of equal length $L$ (metre) from a rigid hook. The whole setup is taken into a satellite where there is no gravity. The two balls are now held by electrostatic forces in a horizontal position. The tension in each string is then: