In a hydrogen-like system,the ratio of the Coulomb force to the gravitational force between an electron and a proton is of the order of:

Three charges $q, Q$ and $+4q$ are placed in a straight line of length $d$ at positions $0, d/3$ and $2d/3$ respectively. In order to make the net force on $q$ be zero, the value of $Q$ should be

$A$ point charge $A$ of $+10 \mu C$ and another point charge $B$ of $+20 \mu C$ are kept $1 \ m$ apart in free space. The electrostatic force on $A$ due to $B$ is $F_1$ and the electrostatic force on $B$ due to $A$ is $F_2$. Then:

The electrical force between two point charges is $200 \ N$. If we increase $10 \ \%$ charge on one of the charges and decrease $10 \ \%$ charge on the other,then the electrical force between them for the same distance becomes . . . . . . . . . . (in $N$)

Three point charges $q$,$-2q$,and $2q$ are placed on the $x$-axis at distances $x=0$,$x=\frac{3}{4}R$,and $x=R$ respectively from the origin as shown. If $q = 2 \times 10^{-6} \, C$ and $R = 2 \, cm$,the magnitude of the net force experienced by the charge $-2q$ is .......... $N$.

$A$ negatively charged oil drop is prevented from falling under gravity by applying a vertical electric field of $100 \ V m^{-1}$. If the mass of the drop is $1.6 \times 10^{-3} \ g$,the number of electrons contained in the drop is: