Two particles of equal mass $m$ and equal charge $q$ are separated by a distance of $16 \text{ cm}$. They do not experience any net force. The value of $\frac{q}{m}$ is . . . . . . (where $G$ is the universal gravitational constant and $\epsilon_0$ is the permittivity of free space).

Write the expression for the Coulombian force acting between two point charges kept in a medium.

$A$ point charge $q_1$ exerts an electric force on a second point charge $q_2$. If a third charge $q_3$ is brought near,the electric force of $q_1$ exerted on $q_2$:

Electrostatic force between two identical charges placed in vacuum at a distance of $r$ is $F$. $A$ slab of width $\frac{r}{5}$ and dielectric constant $9$ is inserted between these two charges. The new force between the charges is:

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 the distance between two equal point charges is doubled,what would happen to the force between them?