The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).

Two unit negative charges are placed on a straight line. $A$ positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium,the value of $q$ (in $C$) is

$A$ charge $Q$ is divided into two charges $q$ and $Q-q$. The value of $q$ such that the force between them is maximum,is

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\overrightarrow{F}$ between the two is (where $K = \frac{1}{4\pi\varepsilon_0}$):

An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4} \; N \, C^{-1}$ (Millikan's oil drop experiment). The density of the oil is $1.26 \; g \, cm^{-3}$. Estimate the radius of the drop. $(g = 9.81 \; m \, s^{-2}; e = 1.60 \times 10^{-19} \; C)$

The ratio of Coulomb's electrostatic force to the gravitational force between an electron and a proton separated by some distance is $2.4 \times 10^{39}$. The ratio of the proportionality constant $K = \frac{1}{4 \pi \varepsilon_0}$ to the gravitational constant $G$ is nearly (Given that the charge of the proton and electron each $= 1.6 \times 10^{-19} \; C$,the mass of the electron $= 9.11 \times 10^{-31} \; kg$,the mass of the proton $= 1.67 \times 10^{-27} \; kg$):